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,
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
From (1) and (2)
=
⇒ OB × OB = OA × OA
⇒ (OB)^{2} = (OA)^{2}
⇒ OA = OB ...(3)
From (1) to (3), we have
Q5: P and Q are points on the 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 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 the 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 the side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produces D at E. Prove that EL = 2 BL.
So: We have parallelogram ABCD in which M is the midpoint 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 cm and
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 × DG But 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.
⇒
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 rightangled at B and D is the midpoint 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 ABC, AC^{2} = BC^{2} + AB^{2}
= (2 BD)^{2} + AB^{2}
= 4 BD^{2} + AB^{2} ...(1)
In the 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}
124 videos457 docs77 tests

1. What are the different types of triangles? 
2. How can we determine the type of triangle if only the angles are given? 
3. How can we find the area of a triangle? 
4. Can we determine the missing angle in a triangle if the other two angles are known? 
5. Can we determine the length of a side in a triangle if the angles and one side length are known? 

Explore Courses for Class 10 exam
