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In the quadrilateral ABCD, ∠ACB = 90°, AB = 20 cm, BC = 12 cm and AD = CD = 17 cm. Find (i) AC (ii) area of quadrilateral ABCD. - Mathematics

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प्रश्न

In the quadrilateral ABCD, ∠ACB = 90°, AB = 20 cm, BC = 12 cm and AD = CD = 17 cm. Find (i) AC (ii) area of quadrilateral ABCD.

योग
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उत्तर

Given:

  • Quadrilateral ABCD with ∠ACB = 90°
  • AB = 20 cm
  • BC = 12 cm
  • AD = CD = 17 cm

We need to find:

  1. AC 
  2. Area of quadrilateral ABCD

Step 1: Find AC using right triangle ABC

Since ∠ACB = 90°, triangle ABC is a right-angled triangle at C.

Use Pythagoras theorem: AB2 = AC2 + BC2 

Substitute known values: 202 = AC2 + 122
400 = AC2 + 144

AC2 = 256

AC = `sqrt(256)`

AC = 16 cm

Step 2: Find the area of quadrilateral ABCD

Quadrilateral ABCD can be divided into two triangles:

Triangle ABC   ...(Right-angled at C)

Triangle ADC   ...(Isosceles triangle with AD = CD = 17 cm)

Area of triangle ABC:

`"Area" = 1/2 xx AC xx BC`

= `1/2 xx 16 xx 12`

= 96 cm2

Find the area of triangle ADC:

Use Heron’s formula: 

`s = (AD + DC + AC)/2`

= `(17 + 17 + 16)/2`

= 25 cm

`Area = sqrt(s(s - AD)(s - DC)(s - AC))`

= `sqrt(25(25 - 17)(25 - 17)(25 - 16))`

= `sqrt(25 xx 8 xx 8 xx 9)`

= `sqrt(25 xx 576)`

= `sqrt(14400)`

= 120 cm2

Total area of quadrilateral ABCD:

= Area(ΔABC) + Area(ΔADC)

= 96 + 120

= 216 cm2

AC = 16 cm

Area of quadrilateral ABCD = 216 cm2

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अध्याय 17: Mensuration - EXERCISE 17B [पृष्ठ २०६]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
अध्याय 17 Mensuration
EXERCISE 17B | Q 18. | पृष्ठ २०६
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