(a) A PQR E A XYZ
(b) A PQR = A ZYX
(c) A PQR = AYXZ
(18) Find the principal that will amou
annum at simple interest in 4 year.
(19) Solve the following.
(a) Add:
9x2 + 4xy and 10xy - 3
(b) Subtract : 7x - 3y + 8 from -4x -
Answers
Answer:
It is known that, expressions with one term is called monomial, expressions with two terms are binomials, expressions with three terms are trinomials and expression with more than three terms are polynomials.
(i) 7x = Monomial
(ii) 5y−7z = Binomial
(iii) 3x3−5x2−11 = Trinomial
(iv) 1−8a−7a2−7a3 = Polynomial
(v) 5m−3 = Binomial
(vi) 4 = Monomial
(vii) 3y2−7y+5 = Trinomial
Step-by-step explanation:
∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)
130º + ∠SRY = 180º
∠SRY = 50º
XY is a straight line. RQ and RS stand on it.
∴ ∠QRX + ∠QRS + ∠SRY = 180º
70º + ∠QRS + 50º = 180º
∠QRS = 180º − 120º = 60º
Video Solution for lines and angles (Page: 104 , Q.No.: 4)
NCERT Solution for Class 9 math - lines and angles 104 , Question 4
Page No 104:
Question 5:
In the given figure, if AB || CD, ∠APQ = 50º and ∠PRD = 127º, find x and y.
ANSWER:
∠APR = ∠PRD (Alternate interior angles)
50º + y = 127º
y = 127º − 50º
y = 77º
Also, ∠APQ = ∠PQR (Alternate interior angles)
50º = x
∴ x = 50º and y = 77º
Video Solution for lines and angles (Page: 104 , Q.No.: 5)
NCERT Solution for Class 9 math - lines and angles 104 , Question 5
Page No 104:
ANSWER:
Let us draw BM ⊥ PQ and CN ⊥ RS.
As PQ || RS,
Therefore, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.
∴∠2 = ∠3 (Alternate interior angles)
However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
∴ ∠1 = ∠2 = ∠3 = ∠4
Also, ∠1 + ∠2 = ∠3 + ∠4
∠ABC = ∠DCB
However, these are alternate interior angles.
∴ AB || CD
Question 1:
In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.
ANSWER:
It is given that,
∠SPR = 135º and ∠PQT = 110º
∠SPR + ∠QPR = 180º (Linear pair angles)
⇒ 135º + ∠QPR = 180º
⇒ ∠QPR = 45º
Also, ∠PQT + ∠PQR = 180º (Linear pair angles)
⇒ 110º + ∠PQR = 180º
⇒ ∠PQR = 70º
As the sum of all interior angles of a triangle is 180º, therefore, for ΔPQR,
∠QPR + ∠PQR + ∠PRQ = 180º
⇒ 45º + 70º + ∠PRQ = 180º
⇒ ∠PRQ = 180º − 115º
⇒ ∠PRQ = 65º
Video Solution for lines and angles (Page: 107 , Q.No.: 1)
NCERT Solution for Class 9 math - lines and angles 107 , Question 1
Page No 107:
Question 2:
In the given figure, ∠X = 62º, ∠XYZ = 54º. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
ANSWER:
As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,
∠X + ∠XYZ + ∠XZY = 180º
62º + 54º + ∠XZY = 180º
∠XZY = 180º − 116º
∠XZY = 64º
∠OZY = = 32º (OZ is the angle bisector of ∠XZY)
Similarly, ∠OYZ = = 27º
Using angle sum property for ΔOYZ, we obtain
∠OYZ + ∠YOZ + ∠OZY = 180º
27º + ∠YOZ + 32º = 180º
∠YOZ = 180º − 59º
∠YOZ = 121º
Video Solution for lines and angles (Page: 107 , Q.No.: 2)
NCERT Solution for Class 9 math - lines and angles 107 , Question 2
Page No 107:
Question 3:
In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.
ANSWER:
AB || DE and AE is a transversal.
∠BAC = ∠CED (Alternate interior angles)
∴ ∠CED = 35º
In ΔCDE,
∠CDE + ∠CED + ∠DCE = 180º (Angle sum property of a triangle)
53º + 35º + ∠DCE = 180º
∠DCE = 180º − 88º
∠DCE = 92º
Video Solution for lines and angles (Page: 107 , Q.No.: 3)
NCERT Solution for Class 9 math - lines and angles 107 , Question 3
Page No 107:
Question 4:
In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT = 95º and ∠TSQ = 75º, find ∠SQT.
ANSWER:
Using angle sum property for ΔPRT, we obtain
∠PRT + ∠RPT + ∠PTR = 180º
40º + 95º + ∠PTR = 180º
∠PTR = 180º − 135º
∠PTR = 45º
∠STQ = ∠PTR = 45º (Vertically opposite angles)
∠STQ = 45º
By using angle sum property for ΔSTQ, we obtain
∠STQ + ∠SQT + ∠QST = 180º
45º + ∠SQT + 75º = 180º
∠SQT = 180º − 120º
∠SQT = 60º
Video Solution for lines and angles (Page: 107 , Q.No.: 4)
NCERT Solution for Class 9 math - lines and angles 107 , Question 4
= 28º and ∠QRT = 65º, then find the values of x and y.
ANSWER:
It is given that PQ || SR and QR is a transversal line.
∠PQR = ∠QRT (Alternate interior angles)
x + 28º = 65º
x = 65º − 28º
x = 37º
By using the angle sum property for ΔSPQ, we obtain
∠SPQ + x + y = 180º
90º + 37º + y = 180º
y = 180º − 127º
y = 53º
x = 37º and y = 53º
ANSWER:
In ΔQTR, ∠TRS is an exterior angle.
∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS − ∠TQR (1)
For ΔPQR, ∠PRS is an external angle.
∠QPR + ∠PQR = ∠PRS
∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)
∠QPR = 2(∠TRS − ∠TQR)
∠QPR = 2∠QTR [By using equation (1)]
∠QTR = ∠QPR