Q 1. The side QR of ∆PQR is produced to a point S. If the bisector of ∆PQR and ∆PRS meet at
point T, then prove that ∠ QTR =
1_
2
∠ QPR
Q 2. L and m are two parallel lines intersected by another pair of parallel lines p and q. show
that ∆ABC ≅ ∆CDA
Q 31. ABC is an isosceles triangle with AB=AC . Draw A ⊥ BC to show that ∠ B = ∠ C
Q 3. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show
that ∠ A >∠ C and ∠ B >∠ D.
Q 4. Construct a triangle ABC in which BC = 7cm, ∠ B =750
and AB + AC = 13 cm.
Q 5. Two coins are teased simultaneously 500 times and we get; no head 120 times.
Two heads : 105 times; One head : 275 times, find probability.
Q 6. Factories : x3
+ 13 x2
+ 32x + 20
Q 7. Factorise : 2x2
+ y2
+8z −¿ 2√2 xy + 4√2 yz– 8 xz
Q 8 . PQR is a line. Ray OR is dperpendicular to line PQ.QS is another ray lying between rays OP
& OR. Prove that ∠ ROS =
1_
2
(∠ QOS −∠ POS
Q 9. Line AB is the side of quadrilateral ABCD in which AD = BC and ∠ DAB = ∠ CBA. Prove
that
(i) ∆ABD ≅ ∆BAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC
Q 10. In right triangle ABC, Right angled at C, M is the mid-point of hypotenuse AB. C is joined
to M and produced to a point D is joined to point B. show that:
(i) ∆AMC ≅ ∆BMD (ii) ∠ DBC is a right angle (iii) ∆DBC ≅ ∆ACB
Cm =
1_
2
AB
Q 11. Represent √ 9.3 on the number line. Write all steps of construction.
NO SPAMS ‼️ PLZ answer all questions according yo 9th class .
and in details.
Answers
QUESTION 1
ANSWER
here
∠QPT = ∠TQR ( LET THEM BE 'X')
∠QRT = ∠TRS ( LET IT BE 'Y')
THEN
IN TRIANGLE PQR
∠QPR + 2X + (180° -2Y) = 180° ∠QPR + 2X - 2Y = 0
∠QPR = 2Y - 2X
∠QPR/2 = Y - X -----(1)
IN TRIANGLE QRT
X + (180° - Y) + ∠QTR = 180°X - Y + ∠QTR = 0∠QTR = Y - X --------(2)
USING 1 AND 2 WE GET∠QPR/2 = ∠QTR
HENCE PROVED
_______________________________________________
question 2 answer
In ΔABC and ΔCDA,
∠BAC = ∠DCA (Alternate interior angles, as p || q)
AC = CA (Common)
∠BCA = ∠DAC (Alternate interior angles, as l || m)
∴ ΔABC ≅ ΔCDA (By ASA congruence rule)
_______________________________________________
1....... :)
=>
Given,
Bisectors of ∠PQR & ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR
(Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒∠QTR=∠TRS–∠TQR — (i)
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]
⇒∠QPR= 2∠TRS – 2∠TQR
⇒∠QPR= 2(∠TRS – ∠TQR)
⇒ 1/2∠QPR = ∠TRS – ∠TQR — (ii)
Equating (i) and (ii)
∠QTR= 1/2∠QPR
Hence proved.
2.......... :)
=>
In ΔABC and ΔCDA,
∠BAC = ∠DCA (Alternate interior angles, as p || q)
AC = CA (Common)
∠BCA = ∠DAC (Alternate interior angles, as l || m)
∴ ΔABC ≅ ΔCDA (By ASA congruence rule)
31.......... ;)
=>
Let the perpendicular bisector from A be AD
Consider ΔABD ↔ ΔACD
1) AD = AD [Common Side]
2) m∠ADB = m∠ADC = 90°
3) BD = CD [AD is the bisector]
∴ By SAS Test of congruence,
ΔABD is congruent to ΔACD
∴m∠B = m∠C
3.......... :)
=>
Given:
In quadrilateral ABCD, AB smallest & CD is longest sides.
To Prove: ∠A>∠C
& ∠B>∠D
Construction: Join AC.
Mark the angles as shown in the figure..
Proof:
In △ABC , AB is the shortest side.
BC > AB
∠2>∠4 …(i)
[Angle opposite to longer side is greater]
In △ADC , CD is the longest side
CD > AD
∠1>∠3 …(ii)
[Angle opposite to longer side is greater]
Adding (i) and (ii), we have
∠2+∠1>∠4+∠3
⇒∠A>∠C
Similarly, by joining BD, we can prove that
∠B>∠D
4.......... :)
=>
Given : Base BC = 7cm, angle B = 75° and sum of two sides AB + AC = 12 cm.
Required : To Construct ∆ABC
STEPS OF CONSTRUCTION :
1. Draw a ray BX and cut off a line segment BC = 7cm; from it.
2. At B; construct angle YBX = 75°
3. With B as centre and radius = 12 cm (because AB + AC = 12cm) draw an arc to meet BY at D.
4. Join CD
5. Draw Perpendicular bisector PQ of CD intersecting BD at A.
6. Join AC
Then ABC is the required triangle.
A lies on perpendicular bisector of CD.
Therefore, AC = AD
=> AB = BD - AD
=> AB = BD - AC
=> AB + AC = BD = 12cm,
5........... ;)
=>
Total number of trials=500
H^2→two heads showing up
H→ one head showing up
T^2→two tails showing up
P(H^2)=105/500=21/100
P(H^1)=275/500=13/25
P(T^2)=120/500=6/25.
6.......... ;)
=> ........Don't know this sry...
8.......... ;)
=>
Given:
OR perpendicular to PQ.
Proof:
/* From the figure ,
/* Adding (1) and (2)
[tex] ∠ROS +∠ROS = ∠QOS - ∠QOR +∠ROP-∠POS
[tex] Since,∠QOR= ∠ROP = 90(given)
]
9......... ;)
=>
in triangle ABD and triangle BAC
AB=BC (GIVEN)
∠DAB = ∠ CBA (GIVEN)
AB = BA (COMMON)
THEREFORE, TRIANGLE ABD ≅ TRIANGLE BAC ( BY SAS)
THEREFORE, BD = AC (CPCT)
ALSO, ∠ ABD = BAC (CPCT)
10.......... ;)
=>
In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.
It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.
Criteria for congruence of triangles:
There are 4 criteria for congruence of triangles.
SAS( side angle side):
Two Triangles are congruent if two sides and the included angle of a triangle are equal to the two sides and included angle of the the other triangle.
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Given:
In right angled ∆ABC,
∠C = 90°,
M is the mid-point of AB i.e, AM=MB & DM = CM.
To Prove:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
Proof:
(i) In ΔAMC & ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Hence, ΔAMC ≅ ΔBMD
( by SAS congruence rule)
ii) since, ΔAMC ≅ ΔBMD
AC=DB. (by CPCT)
∠ACM = ∠BDM (by CPCT)
Hence, AC || BD as alternate interior angles are equal.
Then,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°
Hence, ∠DBC = 90°
(ii) In ΔDBC & ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC ( proved in part ii)
Hence, ΔDBC ≅ ΔACB (by SAS congruence rule)
(iii) DC = AB (ΔDBC ≅ ΔACB)
⇒ DM + CM =AB
[CD=CM+DM]
⇒ CM + CM = AB
[CM= DM (given)]
⇒ 2CM = AB
Hence, CM=1/2AB
11....... ;)
=>
STEPS OF CONSTRUCTION :
1) Draw a line segment AB of length 9.3 units.
2) Extend the line by 1 unit more such that BC=1 unit .
3) Find the midpoint of AC.
4) Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.
5) Draw an arc DE such that BE=BD.
Therefore, BE=
9.3