In figure below, $\mathrm{ABC}$ is a right triangle right angled at $\mathrm{A} . \mathrm{BCED}, \mathrm{ACFG}$ and $\mathrm{ABMN}$ are squares on the sides $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$ respectively. Line segment $\mathrm{AX} \perp \mathrm{DE}$ meets $\mathrm{BC}$ at Y. Show that:(i) $\triangle \mathrm{MBC} \cong \triangle \mathrm{ABD}$(ii) $\operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC})$(iii) $\operatorname{ar}(\mathrm{BYXD})=\operatorname{ar}(\mathrm{ABMN})$(iv) $\triangle \mathrm{FCB} \cong \triangle \mathrm{ACE}$(v) $\operatorname{ar}(\mathrm{CYXE})=2 \operatorname{ar}(\mathrm{FCB})$(vi) $\operatorname{ar}(\mathrm{CYXE})=\operatorname{ar}(\mathrm{ACFG})$(vii) ar $(\mathrm{BCED})=\operatorname{ar}(\mathrm{ABMN})+\operatorname{ar}(\mathrm{ACFG})$"

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Given:

$\mathrm{ABC}$ is a right triangle right angled at $\mathrm{A} . \mathrm{BCED}, \mathrm{ACFG}$ and $\mathrm{ABMN}$ are squares on the sides $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$ respectively. Line segment $\mathrm{AX} \perp \mathrm{DE}$ meets $\mathrm{BC}$ at Y.

To do:

We have to show that

(i) $\triangle \mathrm{MBC} \cong \triangle \mathrm{ABD}$
(ii) $\operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC})$
(iii) $\operatorname{ar}(\mathrm{BYXD})=\operatorname{ar}(\mathrm{ABMN})$
(iv) $\triangle \mathrm{FCB} \cong \triangle \mathrm{ACE}$
(v) $\operatorname{ar}(\mathrm{CYXE})=2 \operatorname{ar}(\mathrm{FCB})$
(vi) $\operatorname{ar}(\mathrm{CYXE})=\operatorname{ar}(\mathrm{ACFG})$
(vii) ar $(\mathrm{BCED})=\operatorname{ar}(\mathrm{ABMN})+\operatorname{ar}(\mathrm{ACFG})$

Solution:

In $\triangle ABD$ and $\triangle MBC$,

$BC = BD$        (Sides of a square are equal)

$MB = AB$

$\angle MBC = 90^o + \angle ABC$

$= \angle DBC + \angle ABC$

$= \angle ABD$

Therefore, by SAS congruency,

$\triangle MBC = \triangle ABD$

(ii) $ar(\triangle MBC) = ar (\triangle ABD)$.......…(i)

$ar(\triangle ABD) = \frac{1}{2}ar (BYXD) …(ii) (Since$\triangle ABD$and rectangle$BYXD$lie on the same base and between the same parallels) From (i) and (ii), we get,$ar (\triangle MBC) = \frac{1}{2}ar (BYXD)$......(iii)$ar (BYXD) = 2 ar (\triangle MBC)$(iii)$ar (\triangle MBC) = \frac{1}{2} ar (ABMN)$…..(iv) (SInce$\triangle MBC$and square$ABMN$lie on the same base$MB$and between the same parallels$MB$and$NC$) From (iii) and (iv), we get,$ar (BYXD) = ar (ABMN)$(iv) In$\triangle ACE$and$\triangle FCS$,$AC = FCCE = BC$(Sides of the square)$\angle FCB = 90^o + \angle ACB= \angle BCE + \angle ACB= \angle ACE$Therefore, by SAS congruency,$\triangle FCB = \triangle ACE$(v)$ar(\triangle ACE) = ar(AFCB)$....…(vi)$ar(\triangle ACE) = \frac{1}{2} ar(CVXE)$(Since both lie on the same base$CE$and between the same parallels$CE$and$AX$) From (vi) and (vii), we get,$ar (\triangle ACE) = \frac{1}{2} ar (CYXE)= ar (\triangle FCB)$.....…(vii)$ar (CYXE) = \frac{1}{2} ar (\triangle FCB)$(vi)$ar(AFCB) = \frac{1}{2} ar (ACFG)$......…(ix) (Since both lie on the same base$CF$and between the same parallels$CF$and$BG$) From (viii) and (ix), we get,$\frac{1}{2} ar (ALFG) = \frac{1}{2} ar (CYXE)ar (ACFG) = ar (CYXE)$(vii)$ar (BCED) = ar (BYXD) + ar (CYXE)= ar (ABMN) + ar (ACFG)\$         [From(iii) and (vi)]

Hence proved.

Updated on 10-Oct-2022 13:46:30