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

00:00 Calculate the area of triangle ABC

00:14 Equal sides in an isosceles triangle

00:17 AD is perpendicular to BC according to the given

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 Let's substitute the value of side BC

00:50 The given ratio between side AD and BC

00:54 Let's substitute appropriate values

01:06 Let's divide 1.2 into factors (1) and (0.2) and solve

01:15 And this is the height AD

01:23 Let's calculate the area of triangle ABC

01:26 We'll use the formula for calculating triangle area

01:29 (Height(AD) multiplied by base(BC)) divided by 2

01:33 Let's substitute appropriate values according to our calculation

01:37 Let's calculate and solve

01:44 And this is the solution to the problem

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

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

than the length of the side $BC$ .

That is:

$AD=1.2\cdot BC$

$\frac{100}{100}+\frac{20}{100}=\frac{120}{100}=1.2$

$AD=1.2\cdot20=24$

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

$AΔ\text{ABC}=\frac{AD\cdot BC}{2}=\frac{24\cdot20}{2}=\frac{480}{2}=240$

240 cm²