**Chapter 15 - Area**

**Practice set 15.1**

**Practice set 15.1**

**1. If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.**

**Given:**

**Base of the parallelogram = 18 cm**

**Height of the parallelogram = 11 cm**

** **

**To find:**

**Area of parallelogram**

** **

**Solution:**

**Area of parallelogram = Base × Height**

**∴ Area of parallelogram = 18 × 11**

**∴ Area of parallelogram = 198 sq.cm**

** **

**Ans:** Area of parallelogram is 198 sq.cm

**2. If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.**

**Given:**

**Area of the parallelogram = 29.6 sq cm**

**Base of the parallelogram = 8 cm**

** **

**To find:**

**Height of the parallelogram**

** **

**Solution:**

**Area of parallelogram = Base × Height**

**∴ 29.6 = 8 × Height**

**∴ \(\large \frac {29.6}{8}\) = Height**

**∴ Height = 3.7 cm**

** **

**Ans:** Height of the parallelogram is 3.7 cm

**3. Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.**

**Given:**

**Area of the parallelogram = 83.2 sq cm**

**Height of the parallelogram = 6.4 cm**

** **

**To find:**

**Base of the parallelogram**

** **

**Solution:**

**Area of parallelogram = Base × Height**

**∴ 83.2 = Base × 6.4**

**∴ \(\large \frac {83.2}{6.4}\) = Base**

**∴ Base = 13 cm**

** **

**Ans:** Base of the parallelogram is 13 cm

**Practice set 15.2**

**Practice set 15.2****1. Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.**

**Given:**

**Length of 1st diagonal = 15 cm**

**Length of 2nd diagonal = 24 cm**

** **

**To find:**

**Area of rhombus**

** **

**Solution:**

**Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals**

**∴ Area of rhombus = \(\large \frac {1}{2}\) × 15 × 24**

**∴ Area of rhombus = 15 × 12**

**∴ Area of rhombus = 180 sq.cm**

** **

**Ans:** Area of rhombus is 180 sq.cm

**2. Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.**

**Given:**

**Length of 1st diagonal = 16.5 cm**

**Length of 2nd diagonal = 14.2 cm**

** **

**To find:**

**Area of rhombus**

** **

**Solution:**

**Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals**

**∴ Area of rhombus = \(\large \frac {1}{2}\) × 16.5 × 14.2**

**∴ Area of rhombus = 16.5 × 7.1**

**∴ Area of rhombus = 117.15 sq.cm**

** **

**Ans:** Area of rhombus is 117.15 sq.cm

**3. If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?**

**Given:**

**Perimeter of rhombus = 100 cm**

**Length of one diagonal = 48 cm**

** **

**To find:**

**Area of rhombus **

** **

**Solution:**

**Let □ ABCD be the rhombus with perimeter 100 cm.**

**Diagonal AC and BD intersect at point M.**

** **

**We know that,**

**Perimeter of a rhombus = 4 × side**

**∴ 100 = 4 × side**

**∴ \(\large \frac {100}{4}\) = Side**

**∴ Side = 25 cm**

**i.e. AB = 25 cm**

** **

**Also,**

**Diagonals of a rhombus are perpendicular bisectors of each other**

** **

**BD = 48 cm …[Given]**

**∴ BM = \(\large \frac {1}{2}\) × BD**

**∴ BM = \(\large \frac {1}{2}\) × 25**

**∴ BM = \(\large \frac {1}{2}\) × 25**

** **

**In ∆ AMB, ∠ AMB = 90⁰**

**∴ ∆ AMB is a right angled triangle**

**∴ AB² = AM² + BM² …[By applying Pythagoras Theorem]**

**∴ 25² = AM² + 24²**

**∴ 625 = AM² + 576**

**∴ 625 – 576 = AM²**

**∴ AM² = 49**

**∴ \( \sqrt {AM²}\) = \( \sqrt {49}\) …[By Taking square root of both sides] **

**∴ AM = 7 cm**

** **

**Now, **

**AC = 2 × AM**

**∴ AC = 2 × 7**

**∴ AC = 14 cm**

**Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals**

**∴ Area of rhombus = \(\large \frac {1}{2}\) × AC × BD**

**∴ Area of rhombus = \(\large \frac {1}{2}\) × 14 × 48**

**∴ Area of rhombus = 14 × 24**

**∴ Area of rhombus = 336 sq.cm**

** **

**Ans:** Area of rhombus is 336 sq.cm

**4*. If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.**

**Given:**

**Length of one diagonal = 30 cm**

**Area of rhombus = 240 sq.cm**

** **

**To find:**

**Perimeter of rhombus **

** **

**Solution:**

**Let □ PQRS be the rhombus.**

**Diagonal PR and QS intersect at point M.**

** **

**Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals**

**∴ 240 = \(\large \frac {1}{2}\) × PR × QS**

**∴ 240 = \(\large \frac {1}{2}\) × 30 × QS**

**∴ 240 = 15 × QS**

**∴ \(\large \frac {240}{15}\) = QS**

**∴ QS = 16 sq.cm**

** **

**We know that,**

**Diagonals of a rhombus are perpendicular bisectors of each other**

** **

**∴ PM = \(\large \frac {1}{2}\) × RP**

**∴ PM = \(\large \frac {1}{2}\) × 30**

**∴ PM = 15 cm**

** **

**and QM = \(\large \frac {1}{2}\) × QS**

**∴ QM = \(\large \frac {1}{2}\) × 16**

**∴ QM = 8 cm**

** **

**In ∆ PMQ, ∠ PMQ = 90⁰**

**∴ ∆ PMQ is a right angled triangle**

**∴ PQ² = PM² + QM² …[By applying Pythagoras Theorem]**

**∴ PQ² = 15² + 8²**

**∴ PQ² = 225 + 64**

**∴ PQ² = 289**

**∴ \( \sqrt {PQ²}\) = \( \sqrt {289}\) …[By Taking square root of both sides] **

**∴ PQ = 17 cm**

** **

**Now,**

**Perimeter of a rhombus = 4 × PQ**

**∴ Perimeter of a rhombus = 4 × 17**

**∴ Perimeter of a rhombus = 68 cm**

** **

**Ans:** Perimeter of a rhombus is 68 cm

**Practice set 15.3**

**Practice set 15.3****1. In □ABCD, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area of □ABCD.**

**Given:**

**In □ABCD, **

**l(AB) = 13 cm**

**l(DC) = 9 cm**

**l(AD) = 8 cm**

** **

**To find:**

**Area of □ABCD**

** **

**Solution:**

**Area of □ABCD = \(\large \frac {1}{2}\) × Sum of parallel sides × Height**

**∴ Area of □ABCD = \(\large \frac {1}{2}\) × (AB + CD) × AD**

**∴ Area of □ABCD = \(\large \frac {1}{2}\) × (13 + 9) × 8**

**∴ Area of □ABCD = 22 × 4**

**∴ Area of □ABCD = 88 sq.cm**

** **

**Ans:** Area of □ABCD is 88 sq.cm

**2. Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.**

**Given:**

**Parallel side 1 = 8.5 cm**

**Parallel side 2 = 11.5 cm**

**Height of the trapezium = 4.2 cm**

** **

**To find:**

**Area of trapezium **

** **

**Solution:**

**Area of trapezium = \(\large \frac {1}{2}\) × Sum of parallel sides × Height**

**∴ Area of trapezium = \(\large \frac {1}{2}\) × (8.5 + 11.5) × 4.2**

**∴ Area of trapezium = 20 × 2.1**

**∴ Area of trapezium = 42 sq.cm**

** **

**Ans:** Area of trapezium is 42 sq.cm

**3*. □PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of □PQRS.**

**Given:**

**□PQRS is an isosceles trapezium,**

**l(PQ) = 7 cm**

**seg PM ⊥ seg SR**

**l(SM) = 3 cm**

**Distance between two parallel sides = 4 cm**

** **

**To find:**

**Area of □PQRS**

** **

**Solution:**

**In ∆ PMS, ∠ PMS = 90⁰**

**∴ ∆ AMB is a right angled triangle**

**∴ PS² = PM² + SM² …[By applying Pythagoras Theorem]**

**∴ PS² = 3² + 4²**

**∴ PS² = 9 + 16**

**∴ PS² = 25**

**∴ \( \sqrt {PS²}\) = \( \sqrt {25}\) …[By Taking square root of both sides] **

**∴ PS = 5 cm**

** **

**PS = QR [∵ □PQRS is an isosceles trapezium]**

**∴ QR = 5 cm**

** **

**Draw seg QM ⊥ seg RS such that point N lies on seg RS**

**∴ QN = PM [Distance between two parallel lines are same.]**

**∴ QN = 4 cm**

** **

**In ∆ QNR, ∠ QNR = 90⁰**

**∴ ∆ QNR is a right angled triangle**

**∴ QR² = QN² + RN² …[By applying Pythagoras Theorem]**

**∴ 5² = 4² + RN²**

**∴ 25 = 16 + RN²**

**∴ 25 – 16 = RN²**

**∴ RN² = 9**

**∴ \( \sqrt {RN²}\) = \( \sqrt {9}\) …[By Taking square root of both sides] **

**∴ RN = 3 cm**

** **

**In □PMNQ,**

**PM || QN [∵ They are ⊥ to the same line]**

**∴ PQ || MN [Parts of parallel sides]**

** **

**∴ □PMNQ is a parallelogram.**

**∴ MN = PQ [Opposite sides of parallelogram are congruent]**

**∴ MN = 7 cm**

** **

**Now,**

**SR = SM + MN + NR**

**∴ SR = 3 + 7 + 3**

**∴ SR = 13 cm**

** **

**Area of □PQRS = \(\large \frac {1}{2}\) × Sum of parallel sides × Height**

**∴ Area of □PQRS = \(\large \frac {1}{2}\) × (PQ + RS) × PM**

**∴ Area of □PQRS = \(\large \frac {1}{2}\) × (7 + 13) × 4**

**∴ Area of □PQRS = 20 × 2**

**∴ Area of □PQRS = 40 sq.cm**

** **

**Ans:** Area of trapezium is 40 sq.cm

**Practice set 15.4**

**Practice set 15.4****1. Sides of a triangle are 45 cm, 39 cm and 42 cm, find its area.**

**Given:**

**a = 45 cm**

**b = 39 cm**

**c = 42 cm**

** **

**To find:**

**Area of triangle**

** **

**Solution:**

**We know that,**

**semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)**

**∴ s = \(\large \frac {45\, +\, 39\, +\, 42}{2}\)**

**∴ s = \(\large \frac {126}{2}\)**

**∴ s = 63**

** **

**Area of triangle = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)**

**∴ Area of triangle = \( \sqrt {63\, (63\, –\, 45)(63\, –\, 39)(63\, –\, 42) }\)**

**∴ Area of triangle = \( \sqrt {63\, ×\, 18\, ×\, 24\, ×\, 21) }\)**

**∴ Area of triangle = \( \sqrt {9\, ×\, 7\, ×\, 2\, ×\, 9\, ×\, 2\, ×\, 4\, ×\, 3\, ×\, 3\, ×\, 7) }\)**

**∴ Area of triangle = \( \sqrt {9²\, ×\, 7²\, ×\, 2²\, ×\, 2²\, ×\, 3²) }\)**

**∴ Area of triangle = 9 × 7 × 2 × 2 ×3**

**∴ Area of triangle = 756 sq.cm**

** **

**Ans:** Area of triangle is 756 sq.cm

**2. Look at the measures shown in the adjacent figure and find the area of □PQRS.**

**Given:**

**PS = 36 m**

**SR = 15 m**

**PQ = 56 m**

**QR = 25 m**

**∠ PSR = 90⁰**

** **

**To find:**

**Area of □PQRS**

** **

**Solution:**

**In ∆ PSR, ∠ PSR = 90⁰**

**∴ ∆ PSR is a right angled triangle**

**∴ PR² = PS² + SR² …[By applying Pythagoras Theorem]**

**∴ PR² = 36² + 15²**

**∴ PR² = 1296 + 225**

**∴ PR² = 1521**

**∴ \( \sqrt {PR²}\) = \( \sqrt {1521}\) …[By Taking square root of both sides] **

**∴ PR = 39 m**

** **

**Area of ∆PSR = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆PSR = \(\large \frac {1}{2}\) × SR × PS**

**∴ Area of ∆PSR = \(\large \frac {1}{2}\) × 15 × 36**

**∴ Area of ∆PSR = 15 × 18**

**∴ Area of ∆PSR = 270 sq.m …(i)**

** **

**For ∆PQR,**

**semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)**

**∴ s = \(\large \frac {39\, +\, 25\, +\, 56}{2}\)**

**∴ s = \(\large \frac {120}{2}\)**

**∴ s = 60**

** **

**Area of ∆PQR = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)**

**∴ Area of ∆PQR = \( \sqrt {60\, (60\, –\, 39)(60\, –\, 25)(60\, –\, 56) }\)**

**∴ Area of ∆PQR = \( \sqrt {60\, ×\, 21\, ×\, 35\, ×\, 4) }\)**

**∴ Area of ∆PQR = \( \sqrt {4\, ×\, 5\, ×\, 3\, ×\, 3\, ×\, 7\, ×\, 7\, ×\, 5\, ×\, 4) }\)**

**∴ Area of ∆PQR = \( \sqrt {4²\, ×\, 5²\, ×\, 3²\, ×\, 7²) }\)**

**∴ Area of ∆PQR = 4 × 5 × 3 × 7**

**∴ Area of ∆PQR = 420 sq.m …(ii)**

** **

**Now,**

**Area of □PQRS = Area of ∆PSR + Area of ∆PQR**

**∴ Area of □PQRS = 270 + 420 …[From (i) and (ii)]**

**∴ Area of □PQRS = 690 sq.m**

** **

**Ans:** Area of □PQRS is 690 sq.m

**3. Some measures are given in the adjacent figure, find the area of □ABCD.**

**Given:**

**AD = 9 m**

**AB = 40 m**

**BE = 13 m**

**DC = 60 m**

**∠ BAD = 90⁰**

**∠ BEC = 90⁰**

** **

**To find:**

**Area of □ABCD**

** **

**Solution:**

**Area of ∆BAD = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆BAD = \(\large \frac {1}{2}\) × AD × AB**

**∴ Area of ∆BAD = \(\large \frac {1}{2}\) × 9 × 40**

**∴ Area of ∆BAD = 9 × 20**

**∴ Area of ∆BAD = 180 sq.m …(i)**

** **

**Area of ∆BCD = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆BCD = \(\large \frac {1}{2}\) × DC × BE**

**∴ Area of ∆BCD = \(\large \frac {1}{2}\) × 60 × 13**

**∴ Area of ∆BCD = 30 × 13**

**∴ Area of ∆BCD = 390 sq.m …(ii)**

** **

**Now,**

**Area of □ABCD = Area of ∆BAD + Area of ∆BCD**

**∴ Area of □ABCD = 180 + 390 …[From (i) and (ii)]**

**∴ Area of □ABCD = 570 sq.m**

** **

**Ans:** Area of □ABCD is 570 sq.m

**Practice set 15.5**

**Practice set 15.5**** Find the areas of given plots. (All measures are in metres.)**

** 1.**

**Given:**

**AP = 30**

**QA = 50**

**AB = 30**

**BC = 30**

**RC = 25**

**CS = 60**

**BT = 30**

** **

**To find:**

**Area of the given plot**

** **

**Solution:**

**Area of ∆PAQ = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆PAQ = \(\large \frac {1}{2}\) × PA × QA**

**∴ Area of ∆PAQ = \(\large \frac {1}{2}\) × 30 × 50**

**∴ Area of ∆PAQ = 30 × 25**

**∴ Area of ∆PAQ = 750 sq.m …(i)**

** **

**Area of ∆RCS = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆RCS = \(\large \frac {1}{2}\) × CS × RC**

**∴ Area of ∆RCS = \(\large \frac {1}{2}\) × 60 × 25**

**∴ Area of ∆RCS = 30 × 25**

**∴ Area of ∆RCS = 750 sq.m …(ii)**

** **

**Area of ∆PTS = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆PTS = \(\large \frac {1}{2}\) × PS × TB**

**∴ Area of ∆PTS = \(\large \frac {1}{2}\) × 150 × 30**

**∴ Area of ∆PTS = 75 × 30**

**∴ Area of ∆PTS = 2250 sq.m …(iii)**

** **

**□QRCA is a trapezium **

**Area of □QRCA = \(\large \frac {1}{2}\) × Sum of parallel sides × Height**

**∴ Area of □QRCA = \(\large \frac {1}{2}\) × (AQ + CR) × AC**

**∴ Area of □QRCA = \(\large \frac {1}{2}\) × (50 + 25) × 60**

**∴ Area of □QRCA = 75 × 30**

**∴ Area of □QRCA = 2250 sq.cm …(iv)**

** **

**Now,**

**Area of the given plot = Area of ∆PAQ + Area of ∆RCS + Area of ∆PTS + Area of □QRCA**

**∴ Area of the given plot = 750 + 750 + 2250 + 2250**

**∴ Area of the given plot = 6000 sq.m**

** **

**Ans:** Area of the given plot is 6000 sq.m

**2. **

**Given:**

**AB = 24**

**BC = 26**

**BE = 30**

**CE = 28**

**FD = 16**

**∠ BAE = 90⁰**

** **

**To find: **

**Area of the given plot**

** **

**Solution:**

**In ∆ BAE, ∠ BAE = 90⁰**

**∴ ∆ BAE is a right angled triangle**

**∴ BE² = BA² + AE² …[By applying Pythagoras Theorem]**

**∴ 30² = 24² + AE²**

**∴ 900 = 576 + AE²**

**∴ 900 – 576 = AE²**

**∴ AE² = 324**

**∴ \( \sqrt {AE²}\) = \( \sqrt {324}\) …[By Taking square root of both sides] **

**∴ AE = 18 m **

** **

**Area of ∆ABE = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆ABE = \(\large \frac {1}{2}\) × AB × AE**

**∴ Area of ∆ABE = \(\large \frac {1}{2}\) × 24 × 18**

**∴ Area of ∆ABE = 24 × 9**

**∴ Area of ∆ABE = 216 sq.m …(i)**

** **

**Area of ∆CDE = \(\large \frac {1}{2}\) × Base × Height**

**∴ Area of ∆CDE = \(\large \frac {1}{2}\) × CE × DF**

**∴ Area of ∆CDE = \(\large \frac {1}{2}\) × 28 × 16**

**∴ Area of ∆CDE = 28 × 8**

**∴ Area of ∆CDE = 224 sq.m …(ii)**

** **

**For ∆BCE,**

**semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)**

**∴ s = \(\large \frac {26\, +\, 30\, +\, 28}{2}\)**

**∴ s = \(\large \frac {84}{2}\)**

**∴ s = 42**

** **

**Area of ∆BCE = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)**

**∴ Area of ∆BCE = \( \sqrt {42\, (42\, –\, 26)(42\, –\, 30)(42\, –\, 28) }\)**

**∴ Area of ∆BCE = \( \sqrt {42\, ×\, 16\, ×\, 12\, ×\, 14) }\)**

**∴ Area of ∆BCE = \( \sqrt {7\, ×\, 6\, ×\, 4\, ×\, 4\, ×\, 6\, ×\, 2\, ×\, 2\, ×\, 7) }\)**

**∴ Area of ∆BCE = \( \sqrt {7²\, ×\, 6²\, ×\, 4²\, ×\, 2²) }\)**

**∴ Area of ∆BCE = 7 × 6 × 4 × 2**

**∴ Area of ∆BCE = 336 sq.m …(iii)**

** **

**Now,**

**Area of the given plot = Area of ∆ABE + Area of ∆CDE + Area of ∆BCE**

**∴ Area of the given plot = 216 + 224 + 336**

**∴ Area of the given plot = 776 sq.m**

** **

**Ans:** Area of the given plot is 776 sq.m

**Practice set 15.6**

**Practice set 15.6****1. Radii of the circles are given below, find their areas.**

**(1) 28 cm **

**Given:**

**Radius of the circle = 28 cm**

** **

**To find:**

**Area of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ Area of the circle = \(\large \frac {22}{7}\) × 28 × 28**

**∴ Area of the circle = 22 × 4 × 28**

**∴ Area of the circle = 2464 sq.cm**

** **

**Ans: **Area of the circle is 2464 sq.cm

**(2) 10.5 cm **

**Given:**

**Radius of the circle = 10.5 cm**

** **

**To find:**

**Area of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ Area of the circle = \(\large \frac {22}{7}\) × 10.5 × 10.5**

**∴ Area of the circle = 22 × 1.5 × 10.5**

**∴ Area of the circle = 346.5 sq.cm**

** **

**Ans: **Area of the circle is 346.5 sq.cm

**(3) 17.5 cm**

**Given:**

**Radius of the circle = 17.5 cm**

** **

**To find:**

**Area of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ Area of the circle = \(\large \frac {22}{7}\) × 17.5 × 17.5**

**∴ Area of the circle = 22 × 2.5 × 17.5**

**∴ Area of the circle = 962.5 sq.cm**

** **

**Ans:** Area of the circle = 962.5 sq.cm

**2. Areas of some circles are given below, find their diameters.**

**(1) 176 sq cm **

**Given:**

**Area of the circle = 176 sq cm**

** **

**To find:**

**Diameter of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ 176 = \(\large \frac {22}{7}\) × r²**

**∴ \(\large \frac {176\, ×\, 7}{22}\) = r²**

**∴ r² = 56 **

**∴ \( \sqrt {r²}\) = \( \sqrt {56}\) …[By Taking square root of both sides] **

**∴ r = \( \sqrt {56}\) m **

** **

**Diameter of the circle = 2 × radius **

**∴ Diameter of the circle = 2 × \( \sqrt {56}\) **

**∴ Diameter of the circle = 2\( \sqrt {56}\) m**

** **

**Ans: **Diameter of the circle = 2\( \sqrt {56}\) m

**(2) 394.24 sq cm **

**Given:**

**Area of the circle = 394.24 sq cm **

** **

**To find:**

**Diameter of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ 394.24 = \(\large \frac {22}{7}\) × r²**

**∴ \(\large \frac {394.24\, ×\, 7}{22}\) = r²**

**∴ r² = 17.92 × 7**

**∴ r² = 125.44**

**∴ \( \sqrt {r²}\) = \( \sqrt {125.44}\) …[By Taking square root of both sides] **

**∴ r = 11.2 m **

** **

**Diameter of the circle = 2 × radius **

**∴ Diameter of the circle = 2 × 11.2**

**∴ Diameter of the circle = 22.4 m**

** **

**Ans: **Diameter of the circle is 22.4 m

**(3) 12474 sq cm**

**Given:**

**Area of the circle = 12474 sq cm**

** **

**To find:**

**Diameter of the circle**

** **

**Solution:**

**Area of the circle = πr²**

**∴ 12474 = \(\large \frac {22}{7}\) × r²**

**∴ \(\large \frac {12474\, ×\, 7}{22}\) = r²**

**∴ r² = 567 × 7**

**∴ r² = 3969**

**∴ \( \sqrt {r²}\) = \( \sqrt {3969}\) …[By Taking square root of both sides] **

**∴ r = 63 m **

** **

**Diameter of the circle = 2 × radius **

**∴ Diameter of the circle = 2 × 63**

**∴ Diameter of the circle = 126 m**

** **

**Ans: **Diameter of the circle = 126 m

**3. Diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.**

**Given:**

**Diameter of the circular garden = 42 m**

**Width of the road = 3.5 m**

** **

**To find:**

**Area of the road**

** **

**Solution:**

**Diameter of the circular garden = 42 m**

**∴ Radius of the circular garden = \(\large \frac {42}{2}\)**

**∴ Radius of the circular garden = 21 m**

** **

**Area of the circular garden = πr²**

**∴ Area of the circular garden = \(\large \frac {22}{7}\) × 21 × 21**

**∴ Area of the circular garden = 22 × 21 × 3**

**∴ Area of the circular garden = 1386 sq.cm**

** **

**Now,**

**Width of the road = 3.5 m**

**Outer Radius = Radius of the circular garden + Width of the road**

**∴ Outer Radius = 21 + 3.5**

**∴ Outer Radius = 24.5**

** **

**Area of the outer circle = πr²**

**∴ Area of the outer circle = \(\large \frac {22}{7}\) × 24.5 × 24.5**

**∴ Area of the outer circle = 22 × 24.5 × 3.5**

**∴ Area of the outer circle = 1886.5 sq.cm**

** **

**Area of the road = Area of the outer circle – Area of the circular garden**

**∴ Area of the road = 1886.5 – 1386**

**∴ Area of the road = 500.5 sq.m**

** **

**Ans: **Area of the road is 500.5 sq.m

**4. Find the area of the circle if its circumference is 88 cm.**

**Given:**

**Circumference of the circle = 88 cm**

** **

**To find:**

**Area of the circle**

** **

**Solution:**

**We know that,**

**Circumference of the circle = 2πr**

**∴ 88 = 2 × \(\large \frac {22}{7}\) × r**

**∴ 88 = \(\large \frac {44}{7}\) × r**

**∴ \(\large \frac {88\, ×\, 7}{44}\) = r**

**∴ r = 2 × 7**

**∴ r = 14 cm**

** **

**Now,**

**Area of the circle = πr²**

**∴ Area of the circle = \(\large \frac {22}{7}\) × 14 × 14**

**∴ Area of the circle = 22 × 14 × 2**

**∴ Area of the circle = 616 sq.cm**

**Ans: **Area of the circle is 616 sq.cm