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The base of an isosceles triangle is 16 cm. If the height on the base is 4 cm shorter than the equal sides, find the sides, height and area of the triangle.

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Question

The base of an isosceles triangle is 16 cm. If the height on the base is 4 cm shorter than the equal sides, find the sides, height and area of the triangle.

Sum
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Solution

Given:

  • Base (b) of the isosceles triangle = 16 cm.
  • Height (h) on base is 4 cm shorter than the equal sides.

Let the length of each equal side be a cm.

Given height h = a – 4.

Step wise calculation:

1. Since the triangle is isosceles, the altitude on the base bisects it, so half of the base is `16/2` = 8 cm.

2. Using Pythagoras theorem in the right triangle formed by the height and half the base: a2 = h2 + 82 

Substitute h = a – 4: a2 = a – 42 + 64

3. Expand and simplify:

a2 = a2 – 8a + 16 + 64 

a2 = a2 – 8a + 80

4. Cancel a2 on both sides:

0 = –8a + 80

8a = 80 

a = 10

5. Find the height:

h = a – 4 

= 10 – 4

= 6 cm

6. Calculate the area of the triangle:

Area = `1/2 xx b xx h`

= `1/2 xx 16 xx 6`

= 48 cm2

Each equal side a = 10 cm

Height h = 6 cm

Area = 48 cm2

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Chapter 17: Mensuration - EXERCISE 17A [Page 201]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 17 Mensuration
EXERCISE 17A | Q 18. | Page 201
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