The figure obtained by joining the mid-points | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $ABCD$ be a rectangle with sides $AB = 8 \, cm$ and $BC = 6 \, cm$.Let $E, F, G,$ and $H$ be the mid-points of sides $AB, BC, CD,$ and $DA$ re

Vedclass · Accepted Answer

a rhombus of area $24 \, cm^{2}$

Vedclass · Answer

(D) Let $ABCD$ be a rectangle with sides $AB = 8 \, cm$ and $BC = 6 \, cm$.Let $E, F, G,$ and $H$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.By joining these mid-points,we obtain a quadrilateral $EFGH$.According to the mid-point theorem,the quadrilateral formed by joining the mid-points of the sides of a rectangle is a rhombus.The diagonals of this rhombus $EFGH$ are equal to the sides of the rectangle,i.e.,$EG = BC = 6 \, cm$ and $HF = AB = 8 \, cm$.The area of the rhombus is given by $\frac{1}{2} 	imes 	ext{product of diagonals} = \frac{1}{2} 	imes EG 	imes HF$.Area $= \frac{1}{2} 	imes 6 \, cm 	imes 8 \, cm = 24 \, cm^{2}$.Thus,the figure obtained is a rhombus of area $24 \, cm^{2}$.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8 \, cm$ and $6 \, cm$ is:

Similar Questions

Observe the given figure. Does the parallelogram $ABCD$ and the triangle $QBC$ lie on the same base and between the same parallels? If yes,write the common base and the two parallels.

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.

$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD$. $AE$ and $DC$ are produced to meet at $F$. Prove that $\operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$.

In $\Delta ABC,$ $P$ and $Q$ are the points of trisection of $BC$ (i.e.,points dividing $BC$ into three equal parts). Prove that,$\operatorname{ar}(ABP) = \operatorname{ar}(APQ) = \operatorname{ar}(AQC) = \frac{1}{3} \operatorname{ar}(ABC).$

In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module