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Andhra Pradesh SSC Class 10 Solutions For Maths – Similar Triangles (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 8 Similar Triangles
Exercise 8.1:
Question 1:
In ∆PQR, ST is a line such that \(\frac{PS}{SQ}=\frac{PT}{TR}\) and also ∠PST = ∠PRQ prove that ∆PQR is an isosceles triangle.
Solution :
AP SSC 10th Class Textbook Solutions
Question 2:
In given figure, \(LM\parallel CB\) and \(LN\parallel CD\) Prove that = \(\frac{AM}{AB}=\frac{AN}{AD}\).
Solution :
Question 3:
In the given figure, \(DE\parallel AC\) and \(DF\parallel AE\) Prove that = \(\frac{BF}{FE}=\frac{BE}{EC}\).
Solution :
Question 4:
In the given figure, \(AB\parallel CD\parallel EF\). given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm. Calculate the values of x and y.
Solution :
Question 5:
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem).
Solution :
Question 6:
Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side. (Using converse of basic proportionality theorem)
Solution :
Question 7:
In the given figure, \(DE\parallel OQ\) and \(DF\parallel OR\). Show that \(EF\parallel QR\).
Solution :
Question 8:
In the adjacent figure, A, B and C are point on OP, OQ and OR respectively such that \(AB\parallel PQ\) and \(AC\parallel PR\). Show that \(BC\parallel QR\).
Solution :
Question 9:
ABCD is a trapezium in which \(AB\parallel DC\) and its diagonals intersect each other at point ‘O ‘. Show that \(\frac{AO}{BO}=\frac{CO}{DO}\).
Solution :
Question 10:
Draw a line segment of length 7.2 and divide it in the ratio 5 : 3. Measure the two parts.
Solution :
We shall take m = 5 and n = 3.
Steps of Construction :
- Draw any ray AX, making an acute angle with AB.
- Locate 8 (= m + n) points A1, A2, A3, A4, A5 ,A6, A7 and A8 on AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5=A6A7=A7A8
- Join BA8.
- Through the point A5 (m = 5), draw a line parallel to A8B (by making an angle equal to ∆AA5B) at A5 intersecting AB at the point C (see Fig). Then, AC: CB = 5: 3.
Diagram:
Exercise 8.2:
Question 1:
In the given figure, ∠ADE = ∠B
- Show that ∆ABC ~ ∆ADE
- If AD = 3.8 cm, AE = 3.6 cm BE = 2.1 cm BC = 4.2 cm Find DE.
Solution :
Question 2:
The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution :
Question 3:
A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6m above the ground, find the length of her shadow after 4 seconds.
Solution :
Question 4:
Given that ∆ABC ~ ∆PQR, CM and RN are respectively the medians of ∆ABC and ∆PQR. Prove that
- ∆AMC ~ ∆PNR
- \(\frac{CM}{RN}=\frac{AB}{PQ}\)
- ∆CMB ~ ∆RNQ
Solution :
Question 5:
Diagonals AC and BD of a trapezium ABCD with \(AB\parallel DC\) intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that \(\frac{OA}{OC}=\frac{OB}{OD}\).
Solution :
Question 6:
AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = Z prove that \(\frac{1}{X}+\frac{1}{Y}=\frac{1}{Z}\).
Solution :
Question 7:
A flag pole 4m tall casts a 6 m. , shadow. At the same time, a nearby building casts a shadow of 24m. How tall is the building ?
Solution :
Question 8:
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆FEG respectively. If ∆ABC ~ ∆FEG then show that
- \(\frac{CD}{GH}=\frac{AC}{FG}\)
- ∆DCB ~ ∆HGE
- ∆DCA ~ ∆HGF
Solution :
Question 9:
AX and DY are altitudes oftwo similar triangles ∆ABC and ∆DEF. Prove that AX : DY = AB : DE.
Solution :
Question 10:
Construct a triangle shadow similar to the given ∆ABC, with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC.
Solution :
Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of ΔABC.
Steps of Construction :
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Locate 5 points (the greater of 5 and 3 in 5/3) B1, B2, B3, B4 and B5 on BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
- Join B3 (the 3rd point, 3 being smaller of 3 and 5 in) to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C’.
- Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’ (see Fig).
Question 11:
Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution :
Question 12:
Construct an Isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution :
Given
An isosceles triangle whose base is 8cm and altitude is 4cm.Scale factor is 1(1/2) = 3/2.
Steps of construction:
- Draw a line segment BC=8cm
- Draw a perpendicular bisector AD of BC.
- Join AB and AC we get an isosceles triangle ΔABC
- Construct an acute angle ∠CBX downwards.
- On BX make three equal parts.
- Join C to B2 and draw a line through B3 parallel to B2C intersecting the line extended line segment BC at C’
- Again draw a parallel line C’A’ to Ac cutting BP at A’
- ∆A’BC’ is the required triangle.
Exercise 8.3:
Question 1:
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Solution :
Question 2:
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Solution :
Question 3:
D, E, F are midpoints of sides BC, CA, AB of ∆ABC. Find the ratio of areas of ∆DEF and ∆ABC.
Solution :
Question 4:
In ∆ABC, \(XY\parallel AC\) and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Solution :
Question 5:
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution :
Question 6:
∆ABC ~ ∆DEF. BC = 3cm EF = 4cm and area of ∆ABC= 54 cm2. Determine the area Of ∆DEF.
Solution :
Question 7:
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP= 1 cm. and BP= 3 cm., AQ = 1.5 cm., CQ = 4.5 cm. Prove that (area of ∆APQ) = \(\frac{1}{16}\) (area of ∆ABC).
Solution :
Question 8:
The areas of two similar triangles are 81cm2 and 49 cm2 respectively. If the attitude of the bigger triangle is 4.5 cm. Find the corresponding attitude Of the smaller triangle.
Solution :
Exercise 8.4:
Question 1:
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution :
ABCD is a rhombus in which diagonals AC and BD intersect at point O.
To prove that:
AB2+BC2+CD2+DA2=AC2+DB2
In ΔAOB; AB2=AO2+BO2
In ΔBOC; BC2=CO2+BO2
In ΔCOD; CD2=DO2+CO2
In ΔAOD; AD2=DO2+AO2
Adding the above four equations, we get;
AB2+BC2+CD2+DA2
= AO2+BO2+ CO2+BO2+ DO2+CO2+ DO2+AO2
Or
AB2+BC2+CD2+DA2
=2(AO2+BO2+ CO2+DO2)
=2(2AO2+BO2) (since AO2=CO2 and BO2=DO2)
=4(AO2+BO2) ……(1)
Now, let us take the sum of squares of diagonals;
AC2+DB2=(AO+CO)2+(DO+BO)2
=(2AO)2+(2DO)2
=4AO2+4BO2 ……(2)
From equations (1) and (2), it is clear;
AB2+BC2+CD2+DA2=AC2+DB2 thus proved.
Question 2:
ABC is aright triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE2 + CD2 = AC2 + DE2.
Solution :
In right angled ΔABE and ΔDBC, we have
AE2=AB2+BE2 …..(1)
DC2=DB2+BC2 …(2)
Adding (1) and (2)
AE2+ DC2
= AB2+BE2+ DB2+BC2
= (AB2+ BC2)+( BE2+ DB2)
=AC2+DE2 [since AB2+BC2=AC2 in right angled
triangle ABC]
Thus proved.
Question 3:
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Solution :
Question 4:
PQR is a triangle right angled at P and M is a point on QR such that \(PM\bot QR\). Show that PM2 = QM ∙ MR.
Solution :
Let ∠MPR=x
In ∆MPR,
∠MRP=180˚-90˚-x
∠MRP=90˚-x
Similarly in ∆MPQ,
∠MPQ=90˚- ∠MPR
=90˚-x
∠MQP=180˚-90˚-(90˚-x)
∠MQP=x
In ∆QMP and ∆PMR
∠MPQ=∠MRP
∠PMQ=∠RMP
∠MQP=∠MPR
∆QMP ∼ ∆PMR [by AAA criteria]
QM/PM=MP/MR
PM2=MR×QM.
Question 5:
ABD is a triangle right angled at A and \(PM\bot QR\) Show that
- AB2 = BC ∙ BD.
- AC2 = BC . DC
- AD2 = BD ∙ CD.
Solution :
Question 6:
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution :
Since the triangle is right angled at C therefore the side AB is the hypotenuse. Let the base of the triangle be AC and the altitude be BC. Applying the Pythagorean theorem
Hyp2 = Base2+ Alt2
AB2 = AC2 + BC2
Since the triangle is isosceles triangle two of the sides shall be equal.
Therefore AC = BC.
Thus AB2 = AC2 + AC2
AB2 = 2AC2
Question 7:
Solution :
Given,
ΔABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB
To prove,
OA2+OB2+OC2-OD2-OE2-OF2=AF2+BD2+CE2
Construction
Joint point O to A, B and C
Proof
(i) ∠AFO=90˚
AO2=AF2+OF2
⇒ AF2= AO2– OF2 (1)
Similarly BD2=BO2-OD2 (2)
CE2=CO2-OE2 (3)
Adding (1), (2) and (3)
AF2+BD2+CE2= OA2+OB2+OC2-OD2-OE2-OF2
(ii) AF2+BD2+CE2= (AO2-OE2)+(BO2-OF2)+(CO2-OD2)
= AE2+CD2+BF2
Question 8:
A wire attached to vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution :
Question 9:
Two poles of heights 6m and 11m stand on a plane ground. If the distance between the feet of the poles is 12m find the distance between their tops.
Solution :
Given
BC=6m, AD=11m, BC=ED.
So, AE=AD-ED=11-6=5m
BE=CD=12m
Find, AB=?
Solution:
Now, In ΔABE, ∠E=90˚
AB2=AE2+BE2
AB2=52+122
AB2=169
AB=13m
The distance between their tops is 13m.
Question 10:
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD2 = 7AB2.
Solution :
Question 11:
In the given figure, ABC is a triangle right angled at B. D and E are ponts on BC trisect it. Prove that 8AE2 = 3AC2 + 5AD
Solution :
Question 12:
ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.
Solution :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 8 Similar Triangles are helpful to complete your math homework.
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