Question

The parallel sides DC and ab of a Trapezium are 25 cm and 10cm respectively its non parallel sides are each equal to 15 cm find the area of the trapezium​

Answer 1

$\textbf{\huge{\underline{Given:}}}$

● Measure of DC = 25 cm

● Measure of AB = 10 cm

● Non-parallel sides = 15 cm

$\textbf{\huge{\underline{To\:Find:}}}$

Area of Trapezium

$\textbf{\huge{\underline{Solution:}}}$

Let ABCD be the given trapezium in which AB = 10 cm, DC = 25 cm, BC = 15 cm and AD = 15 cm

$\rm \: Through \: B, draw \ratio$

$\star \: \sf BF \parallel AD$

$\star \: \sf BE \perp DC$

Now,

$\hookrightarrow \rm \: FC = DC - DF$

$\hookrightarrow \sf \: FB =25 - 10$

$\hookrightarrow \sf \: FB = \large \boxed{ \sf \: 15 \ cm}$

In triangle FCB,FB= BC = 15 cm

So,It is an Isosceles Triangle

Also,BE is perpendicular to FC. So,E is the midpoint of FC

$\sf \: \therefore \: FE = \dfrac{1}{2} \: \times FC$

$\hookrightarrow \sf \: FE = \dfrac{1}{2} \times 15 \: cm$

$\hookrightarrow \: \sf \: FE = \large \boxed{ \sf \: 7.5 \: cm}$

$\star \rm \: In \: right \: angled \: triangle \: BEF$

$\hookrightarrow \sf \: {BF}^{2} = {FE}^{2} + {BE}^{2}$

$\hookrightarrow \sf \: {15}^{2} = {7.5}^{2} + {BE}^{2}$

$\hookrightarrow \sf \: {BE}^{2} = 225 - 56.25$

$\hookrightarrow \sf {BE}^{2} = 168.75$

$\hookrightarrow \sf BE= \sqrt{168.75}$

$\hookrightarrow \sf BE= \large \boxed{ \sf 13 \: cm} \: (approx)$

Now,Area of Trapezium ABCD =

$\hookrightarrow \rm \: \dfrac{1}{2} (AB + DC ) \times BE$

$\hookrightarrow \rm \: \dfrac{1}{2} (25+ 10 ) \times 13 \:$

$\hookrightarrow \rm \: \dfrac{1}{2} \times 35 \times 13 \:$

$\hookrightarrow \large \boxed{ \sf \: 227.5 \: cm^2 } \: \sf \: (approx)$

Answer 2

Answer:

• Reference in the attachment★

$\textbf{\underline{Given:}}$

☆ AB = 10 cm

☆ DC = 25 cm

☆ Non Parallel sides : AC & BD = 15 cm

$\rule{300}{2}$

$\textbf{\underline{To\:Find\::}}$

☆Area of the trapezium.

$\rule{300}{2}$

$\textbf{\underline{Extra\::}}$

Draw BE on DC so that BE is perpendicular to DC. Then draw BH on DC. So E is the midpoint of DH.

So, HE = DE............(i)

CH = AB = 10 cm

☆ DH = DC - CH

DH = (25 - 10) cm

★ DH = 15 cm

$\rule{300}{2}$

$HE =\frac{1}{2}\times DH$

$HE =\frac{1}{2}\times 15\:cm$

★HE = 7.5 cm

$\rule{300}{2}$

DCH is an isosceles triangle because BH = BD = 15 cm.

∆DEH :- In right angled triangle ,

${BH}^{2} = {BE}^{2}+{HE}^{2}$

${15}^{2}={BE}^{2}+{7.5}^{2}$

$225 = {BE}^{2}+56.25$

${BE}^{2}=225-56.25$

${BE}^{2}=168.75$

$BE = \sqrt{168.75}$

★$BE = 12.99 cm$

$\rule{300}{2}$

$Area\:of\:trapezium=\frac{1}{2}\times(AB+DC)\times BE$

$Area =\frac{1}{2}\times ( 10+25)\times 12.99$

$Area =\frac{1}{2}\times 35 \times 12.99$

$Area =\frac{1}{2}\times 454.65$

★$Area =227.325 {cm}^{2}$

$\large\boxed{Area=227.325{cm}^{2}(Approx.)}$

Hope it helps you ♥ ♥ ♥