CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Given: ABCD is a square, where {tex}\angle{/tex}PQR = 90^oand PB = QC = DR
To prove: (i) QB = RC (ii) PQ = QR
(iii) {tex}\angle{/tex}QPR = 45^o
Proof:

Here,
BC = CD … (Sides of square)
CQ = DR … (Given)
BC = BQ + CQ
{tex}\therefore{/tex} CQ = BC − BQ
{tex}\therefore{/tex} DR = BC – BQ .......(1)
Also,
CD = RC + DR
{tex}\therefore{/tex} DR = CD - RC = BC - RC ........(2)
From (1) and (2), we have,
BC - BQ = BC - RC
{tex}\therefore{/tex} BQ = RC
Now in {tex}\triangle{/tex}RCQ and {tex}\triangle{/tex}QBP
we have, PB = QC … (Given)
BQ = RC … [from (i)]
{tex}\angle{/tex}RCQ = {tex}\angle{/tex}QBP … 90^o each
Hence by SAS(Side-Angle-Side) congruence rule,
{tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP
{tex}\therefore{/tex} QR = PQ … (by cpct)
{tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP and QR = PQ … [from (2)]
{tex}\therefore{/tex} In {tex}\triangle{/tex}RPQ,
{tex}\angle{/tex}QPR = {tex}\angle{/tex}QRP ={tex}\frac{1}{2}{/tex}(180^o - 90^o) = {tex}\frac{90}{2}{/tex}= 45^o
{tex}\therefore{/tex} {tex}\angle{/tex}QPR = 45^o

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Nilanjna Singh 17 hours ago

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In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: ( ) ( ) ( ) ( ) i AMC BMD ii DBC is a right angle. iii DBC ACB 1 iv CM AB

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CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 1 week ago

Given: In right triangle ABC, right angled at C. M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.
To Prove:

{tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD
{tex}\angle{/tex}DBC is a right angle
{tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB
CM = {tex}\frac {1} {2}{/tex}AB

Proof:

In {tex}\triangle{/tex}AMC and {tex}\triangle{/tex}BMD
AM = BM ...[As M is the mid-point]
CM = DM ...[Given]
{tex}\angle{/tex}AMC = {tex}\angle{/tex}BMD ...[Vertically opposite angles]
{tex}\therefore{/tex} {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD proved ...[SAS property] ...(1)
{tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]
{tex}\angle{/tex}ACM = {tex}\angle{/tex}BDM ...[c.p.c.t.]
These are alternate interior angles and they are equal.
{tex}\therefore{/tex} AC {tex}\|{/tex} BD
As AC {tex}\|{/tex} BD and transversal BC intersects them
{tex}\therefore{/tex} {tex}\angle{/tex}DBC + {tex}\angle{/tex}ACB = 180° ...[Sum of the consecutive interior angles of the transversal]
{tex}\angle{/tex}DBC + 90° = 180°
{tex}\angle{/tex}DBC = 180° - 90° = 90°
{tex}\angle{/tex}DBC is a right angle proved.
{tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]
{tex}\therefore{/tex} AC = BD ...[c.p.c.t.] ...(2)
In DDBC and DACB
BC = CB ...[Common]
{tex}\angle{/tex}DBC = {tex}\angle{/tex}ACB ...[each = 90°as proved above]
BD = CA ...[From (2)]
{tex}\therefore{/tex} {tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB ...[SAS property]
DDBC {tex}\cong{/tex} DACB ...[As proved in (iii)]
{tex}\therefore{/tex} DC = AB ...[c.p.c.t.]
{tex}\therefore{/tex} 2CM = AB ...[DM = CM = {tex}\frac {1} {2}{/tex} DC]
{tex}\therefore{/tex} CM ={tex}\frac {1} {2}{/tex} AB

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CBSE > Class 09 > Mathematics

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Preeti Dabral 6 days, 23 hours ago

Let a square ABCD in which L, M, N & O are the midpoints.
In {tex}\triangle{/tex}AML and {tex}\triangle{/tex}CNO
AM = CO (AB = DC and M and O are the midpoints)
AL = CN (AD = BC and L and N are the midpoints)
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{tex}\triangle{/tex}AML {tex}\cong{/tex} {tex}\triangle{/tex}CNO
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similarly {tex}\triangle{/tex}MBN {tex}\cong{/tex} {tex}\triangle{/tex}LDO
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similarly in {tex}\triangle{/tex}LDO
{tex}\angle{/tex}DLO = 45°
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