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PLEASE HELP I WILL GIVE BRAINLIEST I REALLY NEED IT
Given: ABCD is a trapezoid,
AD=10, BC=8,
CK- altitude of triangle ACD =30
Find: the area of ABCD


PLEASE HELP I WILL GIVE BRAINLIEST I REALLY NEED IT Given ABCD Is A Trapezoid AD10 BC8 CK Altitude Of Triangle ACD 30 Find The Area Of ABCD class=

Sagot :

To find the area of the trapezoid we need to find the height of the trapezoid.

Trapezoid

A trapezoid is a quadrilateral which is having a pair of opposite sides as parallel and the length of the parallel sides is not equal.

Area of Trapezoid

The area of a trapezoid is given as half of the product of the height(altitude) of the trapezoid and the sum of the length of the parallel sides.

\rm{ Area\ of\ trapezoid = \dfrac{1} {2}\times height \times (Sum\ of the\ parallel\ Sides)

The area of the trapezoid is 54 units².

Given to us :

ABCD is a trapezoid

AD=10, BC = 8,

CK is the altitude altitude

Area of ∆ACD = 30

Area of ∆ACD,

In ∆ACD,

\begin{gathered}\rm { Area\ \triangle ACD = \dfrac{1}{2}\times base\times height\\\\\ \end{gathered}

Substituting the values,

30 = 1/2 * AD × CK

30 = 1/2 * 10 × CK

(30 * 2)/10 = CK

CK = 6 units

Area of Trapezoid ABCD

\rm{ Area\ of\ trapezoid = \dfrac{1} {2}\times height \times (Sum\ of\ the\ parallell Sides)

Area ABCD = [tex] \frac{1}{2} \times ck \times (ad + bc)[/tex]

Area ABCD = [tex] \frac{1}{2} \times 6 \times (10 + 8)[/tex]

Area ABCD = [tex] \frac{1}{2} \times 6 \times (18)[/tex]

Area ABCD = 54 units²

Hence, the area of the trapezoid is 54 units².