Isosceles Triangle Area: Height 20% Greater than Base with DC=10

Question

Triangle ABC is an isosceles triangle AB=AC

AD is the height of the BC

Given DC=10

The length of the height AD is greater by 20% than the length of the side BC.

Calculate the area of the triangle ABC

AAACCCBBBDDD10

Video Solution

Solution Steps

00:00 Calculate the area of the triangle ABC
00:14 Equal sides in an isosceles triangle
00:17 AD is perpendicular to BC according to the given data
00:21 A perpendicular in an isosceles triangle is also a median
00:26 Therefore BD equals DC equals 10
00:36 The base BC equals the sum of its parts (BD+DC)
00:43 Substitute in the value of side BC
00:50 The given ratio between side AD and BC
00:54 Substitute in the relevant values
01:06 Divide 1.2 into factors (1) and (0.2) and proceed to solve
01:15 This is the height AD
01:23 Calculate the area of the triangle ABC
01:26 Apply the formula for calculating the area of a triangle
01:29 (Height(AD) x base(BC)) divided by 2
01:33 Substitute in the relevant values according to our calculation
01:37 Calculate and solve
01:44 This is the solution

Step-by-Step Solution

Given that it is an isosceles triangle if DC=10 DC=10 then BC=20 BC=20 .

We are told that the height AD AD is greater by20 20% percent

than the length of the sideBC BC .

That is:

AD=1.2BC AD=1.2\cdot BC

100100+20100=120100=1.2 \frac{100}{100}+\frac{20}{100}=\frac{120}{100}=1.2

AD=1.220=24 AD=1.2\cdot20=24

From here we can calculate the area of the triangle ΔABC ΔABC :

AΔABC=ADBC2=24202=4802=240 AΔ\text{ABC}=\frac{AD\cdot BC}{2}=\frac{24\cdot20}{2}=\frac{480}{2}=240

Answer

240 cm²