Question

draw a right Triangle in which the side (other than hypotenuse) a length 4 cm and 3 cm then construct another triangle whose sides are 5/3 times the corresponding sides of the given Triangle

Answer 1

Answer:

The new triangle PQR thus formed have sides PQ = $\dfrac{20}{3}$ cm , QR = 5 cm , PR = $\dfrac{25}{3}$ cm

Step-by-step explanation:

Given as :

For A Triangle ABC

The base of right triangle = BC = 3 cm

The perpendicular of right triangle = AB = 4 cm

Let The hypotenuse of triangle = AC = h cm

According to question

Hypotenuse² = perpendicular² + base²

Or, AC² = AB² + BC²

Or, AC² = 4² + 3²

Or, AC² = 25

∴ AC = √25

i,e Hypotenuse = AC = 5 cm

Again

The other right triangle PQR

The sides of Δ PQR =  $\dfrac{5}{3}$ × Δ ABC

So, measure of side PQ = $\dfrac{5}{3}$ × AB

Or, PQ = $\dfrac{5}{3}$ × 4 cm

i.e PQ = $\dfrac{20}{3}$  cm

Again

Measure of side QR = $\dfrac{5}{3}$ × BC

Or, QR = $\dfrac{5}{3}$ × 3 cm

Or, QR = 5 cm

Again

Measure of side PR =  $\dfrac{5}{3}$ × AC

or, PR =  $\dfrac{5}{3}$ × 5

Or, PR =  $\dfrac{25}{3}$ cm

Hence, The new triangle PQR thus formed have sides PQ = $\dfrac{20}{3}$ cm , QR = 5 cm , PR = $\dfrac{25}{3}$ cm  Answer