Question

In ΔABC, AB = 6 cm, BC = 8 cm and ∠ABC = 90°. An isosceles ΔACD is described on the base of AC where AD = CD = 13 cm. Find (i) AC and (ii) Area of ABCD.

Answer 1

Given:

• In ΔABC, AB = 6 cm, BC = 8 cm and ∠ABC = 90°.
• An isosceles triangle ACD is described on base AC, where AD = CD = 13 cm.
• Need to find (i) length AC and (ii) the area of quadrilateral ABCD.

Stepwise Calculation:

i. Finding AC:

Since ∠ABC = 90°, triangle ABC is right angled at B.

Using Pythagoras theorem for right-angled triangle ABC,

`AC = sqrt(AB^2 + BC^2)` 

= `sqrt(6^2 + 8^2)` 

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10 cm

ii. Finding the area of ABCD:

Area of ABCD = Area of ΔABC + Area of ΔACD

Area of ΔABC: Since ABC is right angled at B

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

= `1/2 xx 6 xx 8`

= 24 cm²

Area of ΔACD isosceles with AD = CD = 13 cm:

Use Heron’s formula to find the area.

Sides are AC = 10 cm, AD = 13 cm, CD = 13 cm.

Semi-perimeter, `s = (AC + AD + CD)/2` 

= `(10 + 13 + 13)/2` 

= `36/2` 

= 18 cm.

`"Area" = sqrt[s(s - AC)(s - AD)(s - CD))` 

= `sqrt(18(18 - 10)(18 - 13)(18 - 13))` 

= `sqrt(18 xx 8 xx 5 xx 5)` 

= `sqrt(3600)` 

= 60 cm²

Therefore, total area of ABCD

= 24 + 60 

= 84 cm²

Thus, AC = 10 cm and area of quadrilateral ABCD = 84 cm².