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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
In \(\angle ACD={ 90 }^{ 0 }\) and \(CD\bot AB\) Prove that \(\cfrac { { BC }^{ 2 } }{ { AC }^{ 2 } } =\cfrac { BD }{ AD } \)
2.
In figure 0 is any point inside a rectangle ABCD. Prove that OB2 + OD2 = OA2 + OC2
3.
Prove that in a right triangle, the square of 8. the hypotenuse is equal to the sum of the squares of the others two sides.
4.
BL and CM are medians of a triangle ABC right angled at A.
Prove that 4(BL2 + CM2) = 5BC2.
5.
In \(AD\bot BC\) prove that AB2 + CD2 = BD2 + AC2.
1.
\(\Delta ACD\sim \Delta ABC\)
So,
\(\cfrac { AC }{ AB } =\cfrac { AD }{ AC } \)
AC2 = AB ·AD
Similarly \(\Delta BCD\sim \Delta BAC\)
So,
\(\cfrac { BC }{ BA } =\cfrac { BD }{ BC } \)
BC2 = BA·BD
From (1) and (2)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)
2.
Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB2 = BP2 + OP2
Similarly from \(\Delta OQD\)
OD2 = OQ2 + DQ2
From \(\Delta OQC\)
OC2 = OQ2 + CQ2
\(\Delta OAP\) we have
OA2 = AP2 +OP2
Adding (1) and (2)
OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2
(As BP = CQ and DQ = AP)
= CQ2 + OP2 + OQ2 + AP2
= CQ2 + OQ2 + OP2 + AP2
= OC2+ OA
[From (3) and (4)]
3.
We are given a right triangle ABC right angled at B.
We need to prove that AC2 = AB2 + BC2
Let us draw \(BD\bot AC\)
Now,\(\Delta ADB\sim \Delta ABC\)
\(\cfrac { AD }{ DB } =\cfrac { BC }{ AC } \)
(sides are proportional)
Also,
\(\Delta BDC\sim \Delta ABC\)
\(\cfrac { CD }{ BC } =\cfrac { BC }{ AC } \)
CD·AC = BC2 ..(2)
Adding (1) and (2)
AD .AC + CD . AC = AB2+ BC2
AC(AD + CD) = AB2 + BC2
AC.AC = AB2 + BC2
AC = AB2 + BC2
4.
BL and CM are medians at the \(\triangle\)ABC in which
\(A=\angle { 90 }^{ 0 }\)
From \(\triangle\)ABC
BC2 = AB2 + AC2
(Pythagoras theorem)
From \(\Delta ABL\)
BL2 = AL2 + AB2
\({ BL }^{ 2 }=\left( \cfrac { { AC }^{ 2 } }{ 2 } \right) +{ AB }^{ 2 }\)
(L is the mid-point at AC)
\({ BL }^{ 2 }=\cfrac { { AC }^{ 2 } }{ 4 } +{ AB }^{ 2 }\)
4BL2 = AC2 + 4AB2
From \(\Delta CMA\)
CM2 = AC2 + AM2
\({ CM }^{ 2 }={ Ac }^{ 2 }+\left( \cfrac { AB }{ 2 } \right) ^{ 2 }\)
(M is the mid-point at AB)
\({ CM }^{ 2 }={ AC }^{ 2 }+\cfrac { { AB }^{ 2 } }{ 4 } \)
4CM2 = 4AC2+ AB2
Adding (2) and (3), we have
4(BL2 + CM2) = 5(AC2 + AB2)
4(BL2 + CM2) = 5BC2
5.
From \(\Delta ADC\) we have
AC2 = AD2 + CD2 ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB2 = AD2 + BD2 ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB2 - AC2 = BD2 - CD2
AB2 + CD2 = BD2 + AC2
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