Question

Find the area of the triangle if A and B are squares with sides that are 5 cm and 8cm respectively.

Answer 1

Answer:

12.5 cm^2

Step-by-step explanation:

This problem looks hard, but it's much easier than it seems.

The calculations are very simple, and the Pythagorean theorem is not needed.

The area of the large triangle is the sum of the areas of the two squares minus the areas of three triangles: above left on 5-cm square, below the triangle being part of the 5-cm square and part of the 8-cm square, and above left of the 8-cm square.

sum of area of squares = (8 cm^2) + (5 cm)^2 = 64 cm^2 + 25 cm^2 = 89 cm^2

area of triangle on top left of 5-cm square:

a = bh/2 = (5 cm)(5 cm)/2 = 12.5 cm^2

area of triangle on bottom of both squares and part in each square:

a = bh/2 = (5 cm + 8 cm)(8 cm)/2 = 52 cm^2

area of triangle on top left of 8-cm square:

a = bh/2 = (8 cm)(8 cm - 5 cm)/2 = 12 cm^2

area of large triangle =

= 89 cm^2 - 12.5 cm^2 - 52 cm^2 - 12 cm^2

= 12.5 cm^2

Answer 2

Answer:

A = 12.5cm²

Step-by-step explanation:

Look at the picture.

We have two triangles Δ₁ and Δ₂.

The formula of an area of a triangle is:

$A=\dfrac{ah}{2}$

b - base

h - height

The heights of triangles are equal to sides 5cm and 8cm. We need the length of a common base of triangles.

The vertical sides of a squares are parallel. Therefore 5cm, (5+8)cm, b and 8cm are in proportion:

$\dfrac{b}{5}=\dfrac{8}{5+8}$

$\dfrac{b}{5}=\dfrac{8}{13}$         multiply both sides by 5

$b=\dfrac{40}{13}\ (cm)$

We know: a + b = 5.

Therefore:

$a+\dfrac{40}{13}=5$

$a+\dfrac{40}{13}=\dfrac{65}{13}$        subtract 40/13 from both sides

$a=\dfrac{25}{13}$

Calculate the area of the triangle:

$A=\dfrac{1}{2}\cdot\dfrac{25}{13}\cdot5+\dfrac{1}{2}\cdot\dfrac{25}{13}\cdot8$      distribute

$A=\dfrac{1}{2}\cdot\dfrac{25}{13}(5+8)=\dfrac{1}{2}\cdot\dfrac{25}{13}\cdot13$    cancel 13

$A=\dfrac{1}{2}\cdot25=\dfrac{25}{2}=12.5\ (cm^2)$

You can also use the proportion resulting from the similarity of the right triangles.