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Q1.If one diagonal of a trapezium divides the other diagonal in the ratio of 1: 2. Prove that one of the parallel sides is double the other.
Sol. Since ABCD is a trapezium,
∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
∴ Δ APB ~ Δ CPD
Q2. In the figure, ABC is a D such that BD ⊥ AC and CE ⊥ AB. Prove that:
PD × BP = PC × EP
Sol. In Δ BEP and Δ CPD, we have:
<BPE = <CPD
[Vertically opp. angles]
<BEP = <CDP [Each = 90°]
∴Using AA similarity, we have
Δ BEP ~ Δ CDP
∴Their corresponding sides are proportional,
⇒
⇒ BP × PD = EP × PC
[By cross multiplication]
Q3. AB is a line segment. PB ⊥ AB and QA ⊥ AB are such that PO = 5 cm and QO = 8 cm. If ar (D POB) = 250 cm2, then find the area of D QOA.
Sol. In Δ QOA and Δ POB,
<QOA = <BOP
[Vertically opposite angles]
<QAO = <PBO [Each = 90°]
∴Using AA similarity, we have:
Q4. In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.
Sol. We have a trapezium ABCD in which AB y DC.
Since Δ BOC ~ Δ AOD [Given]
.(1)
In Δ ODC and ΔOBA,
<COD = <AOB
[Vertically opp. angles]
<ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
Δ ODC ~ Δ OBA
...(2)
From (1) and (2)
=
⇒ OB × OB = OA × OA
⇒ OB2 = OA2 ⇒ OA = OB ...(3)
From (1) to (3) we have
Q5. P and Q are points on sides of AB and AC respectively of Δ ABC. If AP = 3 cm, PB= 9 cm, AQ = 5 cm and QC = 15 cm, then show that BC = 4 PQ.
Sol. We have Δ ABC in which P and Q are such that
AP = 3 cm, PB = 9 cm
AQ = 5 cm, QC = 15 cm
i.e., PQ divides AB and AC in the same ratio
∴ PQ y BC
Now, in Δ APQ and Δ ABC
<P = <B [Corresponding angles]
<A = <A [Common]
⇒ Using AA similarity,
Δ APQ ~ Δ ABC
⇒ =
[Œ AB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
Q6. On one of the longer sides PQ of a rectangle PQRS, a point O is taken such that
SO^{2} = PO· PQ
Prove that: Δ POS ~ Δ OSR.
Sol.
We have a rectangle PQRS such that
SO^{2} = PO· PQ
i.e., SO × SO = PO × PQ
⇒ = ...(1)
[Œ PQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:
<1 = <2 [OE PQ y SR,
opp. sides of a rectangle]
⇒ Using SAS similarity, we have
Δ POS ~ Δ OSR
Q7. Determine the length of the altitude AD of an isosceles D ABC in which AB = AC = 2a cm and BC = a cm.
Sol. We have Δ ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.
In Δ ADB and Δ ADC
<ADB = <ADC [Each = 90°]
<B = <C [Opp. angles to equal sides of a D]
∴Δ ADB ~ Δ ADC
=
⇒
Now in right Δ ABD, we have
AB^{2} = AD^{2} + BD^{2}
⇒ AD^{2} = AB^{2}  BD^{2}
= (AB + BD) (AB  BD)
=
Q8. In an equilateral triangle with side ‘a’, prove that its area .
Sol. We have Δ ABC such that
AB = BC = AC = a
Let us draw altitude AD ⊥ BC.
In Δ ADC and Δ ADB
AD = AD [Common]
AC = AB [Each = a]
<ADC = <ADB [Each = 90°]
∴Δ ADC ≌ Δ ADB [RHS congruency]
∴DC = DB =
Now, in right Δ ADB,
AB^{2} = AD^{2} + DB^{2}
⇒ AD^{2} = AB^{2}  DB^{2}
= (AB + DB) (AB  DB)
=
Now, area of Δ ABC = 1/2× Base × altitude
Thus, the area of an equilateral triangle
=
Q9. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Sol.We have Δ ABC in which AB = AC = CA and an altitude AD ⊥ BC.
In Δ ADB and Δ ADC
<ADB = <ADC [Each = 90°]
AB = AB [Given]
AD = AD [Common]
Using RHS congruency, we have
< ADB ≌ < ADC
⇒ DB = DC = ...(1)
Now, in right Δ ADB, we have:
AB^{2} = AD^{2} + BD^{2} [Using Pythagoras Theorem]
⇒ 3 [Side of the equilateral triangle]
= 4 [Altitude]2.
Q10. ABC is a right triangle in which <C = 90° and CD ⊥ AB. If BC = a, CA = b, AB = c and CD = p, then prove that:
=
Sol. We have a right Δ ABC such that <C = 90°.
Also, CD ⊥ AB
Now, ar (Δ ABC) =1/2 × Base × Height
= ...(1)
Also,
. .. (2)
From (1) and (2), we have
=
Dividing throughout by abp, we have:
=
Squaring both sides,
= = ...(3)
Now, In right D ABC,
AB^{2} = AC^{2} + BC^{2}
⇒ c^{2} = b^{2} + a^{2} ...(4)
∴ From (3) and (4), we get
=
Q11. ABC is a right triangle, rightangled at A, and D is the midpoint of AB. Prove that
BC^{2} = CD^{2} + 3 BD^{2}.
Sol. We have a right Δ ABC in which <A = 90°
∴Using Pythagoras Theorem, we have:
BC^{2} = AB^{2} + AC^{2} ...(1)
Again, Δ ACD is right D, <A = 90°
∴CD^{2} = AD^{2} + AC^{2}
...(2) [Using Pythagoras Theorem]
Subtracting (2) from (1), we get
BC^{2}  CD^{2} = AB^{2}  AD^{2} ...(3)
Since D is the midpoint of AB
∴2 BD = AB and AD = BD ...(4)
From (3) and (4), we have:
∴BC^{2}  CD^{2} = (2 BD)^{2}  (BD)^{2}
= 4 BD^{2}  BD^{2}
BC^{2} = CD^{2} + 3 BD^{2}
Q12. In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC.
Sol. Let us draw EOF y AB ⇒ OE ⊥ AD and OF ⊥ BC
In Δ OFB
Œ <F = 90°
∴Using Pythagoras theorem, we have:
CB^{2} = OF^{2} + BF^{2} ...(1)
In Δ OED
<E = 90°
∴Using Pythagoras theorem, we have:
OD^{2} = OE^{2} + DE^{2} ...(2)
Adding (1) and (2), we get
OB^{2} + OD^{2} = OF^{2} + BF^{2} + OE^{2} + DE^{2}
= OF^{2} + AE^{2} + OE^{2} + CF^{2}
[Œ BF = AE and CF = DE]
= (OF^{2} + CF^{2}) + (OE^{2} + AE^{2})
= OC^{2} + OA^{2}
= OC^{2} + 52
⇒ 62 + 82 = OC^{2} + 52
⇒ 36 + 64 = OC^{2} + 25
⇒ OC^{2} = 36 + 64  25 = 75
⇒ Thus OC = 5√3cm.
Q13. In the figure, if AD ⊥ Bc, then prove that:
AB^{2} + CD^{2} = AC^{2} + BD^{2}
Sol. In D aDc, <ADc = 90°
∴ AD^{2} = AC^{2}  CD^{2} .....(1) (Using Pythagoras Theorem)
Similarly, in D AbD,
⇒ AD^{2} = AB^{2}  DB^{2}.....(2)
From (1) and (2), we have
AB^{2}  DB^{2} = AC^{2}  CD^{2}
⇒ AB^{2} + CD^{2} = AC^{2} + BD^{2}
Q14. In the given figure, AD ⊥ BC and BD = CD. Prove that:
Sol. BD = 1/2 CD
∴ 3 BD = CD
Since BD + DC = BC
∴ BD + 3 BD = BC
⇒ 4 BD = BC
⇒ BD = 1/4 BC
⇒ CD = 3/4 BC
Now, in right Δ ADC, < D = 90°
By Pythagoras theorem, we get
CA^{2} = AD^{2} + CD^{2} ...(1)
Also in right Δ ADB
AD^{2} = AB^{2}  BD^{2} ...(2)
From (1) and (2),
CA^{2} = AB^{2}  BD^{2}+ CD^{2}
Q15. In the given figure, M is the midpoint of side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produce D at E. Prove that EL = 2 BL.
Sol. We have parallelogram ABCD in which M is the mid point of CD.
In Δ EMD and Δ BMC
MD = MC [Œ M is midpoint of CD]
<EMD = <CMB
[Vertically opposite angles]
<MED = <MBC
[Alternate interior angles]
∴ Δ BMC ≌ Δ EMD [AAS congruency]
⇒ BC = ED ⇒ AD = ED ...(1)
[Œ BC = AD, opposite sides of parallelogram]
Now, in Δ AEL and Δ CBL
<AEL = <CBL [Alternate interior angles]
<ALE = <CLB
[Vertically opposite angles]
∴ By AA similarity, we have:
Q16. In the given figure, Δ ABC is rightangled at C and DE ^ AB. Prove that Δ ABC ~ Δ ADE and hence find the length of AE and DE.
Sol.In Δ ABC and Δ ADE, we have:<A = <A [Common]
<C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
...(1)In right Δ ABC, <C = 90°Using Pythagoras theorem, we have:
AB^{2} = BC^{2} + AC^{2}
= 122 + 52
= 144 + 25 = 169
⇒AB = √169 = 13 cm
Now, from (1), we get
⇒ DE = = 2.77 cmAnd = ⇒ AE
Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE^{2} = BD × EC.
Sol. In Δ DBG and Δ ECF
<3 + <1 = 90° = <3 + <4
∴<3 + <1 = <3 + <4
⇒ <1 = <4
<D = E = 90°
∴Using AA similarity, we have:
= BD × EC = EF × DGBut DG = EF = DE
∴BD × EC = DE × DE
⇒ BD × EC = DE^{2}
Thus, DE^{2} = BD × EC
Q18. In the figure, AD ⊥ BC and BD = CD. Prove that 2 CA^{2} = 2 AB^{2} + BC^{2}.
Sol. Œ BD = 1/3 CD
⇒ 3 BD = CD
∴BC = BD + DC
⇒ BC = BD + 3 BD
⇒ BC = 4 BD ...(1)
And From (2) ...(4)In right Δ ADC,
Using Pythagoras theorem,
CA^{2} = AD^{2} + DC^{2}
= From (3) ...(4)In right Δ ADB
Using Pythagoras theorem,
AD^{2} = AB^{2}  BD^{2}
= Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Sol. We have a quadrilateral ABCD such that its diagonals intersect at O and
= <AOB = <COD [Vertically opposite angles]∴ Using SAS similarity, we have
Δ AOB ~ Δ COD
⇒ Their corresponding angles are equal.
i.e., <1 = <2
But they form a pair of int. alt. angles.
⇒ AB y DC
⇒ ABCD is a trapezium.
Q20. Two triangles ABC and DBC are on the same base BC and on the same side of BC in which
<A = <D = 90°. If CA and BD meet each other at E, show that
AE· EC = BE· ED
Sol. We have right Δ ABC and right Δ DBC on the same base BC such that
<A = <D = 90°
In Δ ABE and Δ DCE
<A = <D = 90°
<1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
⇒ ⇒ AE· EC = BE· ED
Q21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
Δ ABE ~ Δ CFB
Sol. We have parallelogram ABCD
In Δ ABE and Δ CFB, we have
<A = <C
[Opposite angles of parallelogram]
<AEB = <EBC
[Alternate angles, AD y BC]
∴ Using AA similarity, we get
Δ ABE ~ Δ CFB
Q22. In Δ ABC, if AD is the median, then show that AB^{2} + AC^{2} = 2 [AD^{2} + BD^{2}].
Sol. AD is a median,
∴ BD = DC
Let us draw AE ≌ BC
Now, in rt. Δs AEB and AEC, we have
AB^{2} = BE^{2} + AE^{2} ...(1)
AC^{2} = CE^{2} + AE^{2} ...(2)
Adding (1) and (2),
AB^{2} + AC^{2} = BE^{2} + AE^{2} + CE^{2} + AE^{2}
= (BD  ED)2 + AE2 +
(CD + DE)2 + AE2
= 2AE^{2} + 2ED^{2} + BD2 + CD^{2}
= 2 [AE^{2} + ED^{2}] + BD^{2} + BD^{2}
[BD = CD]
= 2 [AD]2 + 2BD^{2}
[AE^{2} + ED^{2} = AD^{2}]
= 2 [AD^{2} + BD^{2}]
Thus, AB^{2} + AC^{2} = 2 [AD^{2} + BD^{2}]
Q23. Triangle ABC is right angled at B and D is the mid point of BC. Prove that:
AC^{2} = 4 AD^{2 } 3 AB^{2}
Sol. D is the midpoint of BC.
∴ BC = 2 BD
Now, in Δ ABCAC^{2} = BC^{2} + AB^{2}
= (2 BD)^{2} + AB^{2}
= 4 BD^{2} + AB^{2} ...(1)
In right Δ ABD
Using Pythagoras theorem,
AD^{2} = AB^{2} + BD^{2}
⇒ BD^{2} = AD^{2}  AB^{2} ...(2)
From (1) and (2), we get
AC^{2} = 4 [AD^{2}  AB^{2}] + AB^{2}
⇒ AC^{2} =  4 AB^{2} + 4 AD^{2} + AB^{2}
⇒ AC^{2} =  3 AB^{2} + 4 AD^{2}
or AC^{2} = 4 AD^{2}  3 AB^{2}
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