Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15C
These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C.
Other Exercises
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E
Question 1.
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq.cm and 128 sq.cm. Find the ratio between the lengths of their corresponding sides.
Solution:
Question 2.
A line PQ is drawn parallel to the base BC, of ΔABC which meets sides AB and AC at points P and Q respectively. If AP = \(\frac { 1 }{ 3 }\) PB; find the value of:
Solution:
Question 3.
The perimeters of two similar triangles are 30 cm and 24 cm. If one side of first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution:
Let we are given ΔABC and ΔPQR are similar.
Perimeter of ΔABC = 30 cm.
and perimeter of ΔPQR = 24 cm.
and side BC = 12 cm.
Now we have to find the length of QR, the corresponding side of ΔPQR
ΔABC ~ ΔPQR
Question 4.
In the given figure AX : XB = 3 : 5
Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
Solution:
We are given in the ΔABC, AX : XB = 3 : 5
XY || BC.
Let AX = 3x and XB = 5x
AB = 3x + 5x = 8x.
Now in ΔAXY and ΔABC,
∠AXY = ∠ABC (corresponding angles)
∠A = ∠A (common)
ΔAXY ~ ΔABC (AA Postulate)
Question 5.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given- In ΔABC, PQ || BC in such away that area APQ = area PQCB
To Find- The ratio of BP : AB
Solution:
In ΔABC, PQ || BC.
Z APQ = Z ABC (corresponding angles)
Now in ΔAPQ and ΔABC,
∠APQ = ∠ABC (proved)
∠A = ∠A (common)
ΔAPQ ~ ΔABC (AA postulate)
Question 6.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4
Solution:
In ΔPQR, LM || QR in such away that PM : MR = 3 : 4
(i) In ΔPQR, LM || QR
Question 7.
The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel: PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.
Calculate-
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution:
Question 8.
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm². Calculate
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]
Solution:
(i) Join QC.
In ΔBPQ and ΔCPD,
∠DPC = ∠BPQ (vertically opposite angles.)
∠PDC = ∠BQP (Alternate angles.)
ΔBPQ ~ ΔCDP (AA postulate)
⇒ area ΔCDP = 4 (area ΔBPQ)
⇒ 2 (2 area ΔBPQ) = 2 x 20 = 40 cm² (2 area A BPQ = 20 cm²)
(ii) Area || gm ABCD = area ΔCPD + area ΔADQ – area ΔBPQ
= 40 + 9 (area BPQ) – area BPQ [(AD = CB = 3 BP)]
= 40 + 8 (area ΔBPQ)
= 40 + 8 (10) cm²
= 40 + 80
= 120 cm²
Question 9.
In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm². Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also. Find the area of triangle BCD.
Solution:
In ΔADE, BC || DE
Area of ΔABC = 25 cm²
and area of trapezium BCED = 24 cm²
Area of ΔADE = 25 + 24 = 49 cm²
DE = 14 cm,
Let BC = x cm.
Now in ΔABC and ΔADE,
∠ABC = ∠ADE (corresponding angles)
∠A = ∠A (common)
ΔABC ~ ΔADE (AA postulate)
Question 10.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.
Find-
(i) ΔAPB : ΔCPB
(ii) ΔDPC : ΔAPB
(iii) ΔADP : ΔAPB
(iv) ΔAPB : ΔADB
Solution:
AP : CP = 3 : 5 ⇒ \(\frac { AP }{ CP }\) = \(\frac { 3 }{ 5 }\)
(i) Now in ΔAPB and ΔCPB,
These triangles have same vertex and their bases are in the same straight line
area ΔAPB : area ΔCPB = AP : PC = 3 : 5 or ΔAPB : ΔCPB = 3 : 5
(ii) In ΔAPB and ΔDPC,
∠APB = ∠DPC (vertically opposite angles)
∠PAB = ∠PCD (alternate angles)
ΔAPB ~ ΔDPC (AA postulate)
⇒ area ΔDPC : area ΔAPB = 25 : 9 or ΔDPC : ΔAPB = 25 : 9
(iii) In ΔADP and ΔAPB,
There have the same vertex and their bases arc in the same straight line.
area ΔADP : area ΔAPB = DP : PB
But PC : AP = 5 : 3
ΔADP : ΔAPB = 5 : 3
(iv) Similarly area ΔAPB : area ΔADB = PB : DB = 3 : (3 + 5) = 3 : 8
Question 11.
In the given figure, ARC is a triangle. DE is parallel to BC and \(\frac { AD }{ DB }\) = \(\frac { 3 }{ 2 }\).
(i) Determine the ratios \(\frac { AD }{ AB }\) , \(\frac { DE }{ BC }\)
(ii) Prove that ΔDEF is similar to ΔCBF.
Hence, find \(\frac { EF }{ FB }\)
(iii) What is the ratio of the areas of ΔDEF and ΔBFC?
Solution:
Question 12.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ΔABC and ΔDEC.
Solution:
In the figure DE = 7.8 cm, AB = 10.4 cm
∠ACD = ∠BCE (given)
Adding ∠DCB both sides,
∠ACD + ∠DCB = ∠DCB + ∠BCE
∠ACB = ∠DCE
Now in ΔABC and ΔDCE
∠B = ∠E (given)
∠ACB = ∠DCE (proved)
ΔABC ~ ΔDCE (AA axiom)
Question 13.
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:
Solution:
In the figure, ΔABC is an isosceles triangle in which
AB = AC = 13 cm, BC = 10 cm,
AD ⊥ BC, CE = 8 cm and EF ⊥ AB
(i) Now in ΔADC and ΔFEB
∠C = ∠B (AB = AC)
∠ADC = ∠EFB (each = 90°)
ΔADC ~ ΔFEB (AA axiom)
Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C are helpful to complete your math homework.
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