Constructions
Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle?
Steps of construction: Define a boundary portion \[AB\text{ }=\text{ }5\text{ }cm.\] Develop a right point \[SAB\]at point \[A.\] Draw a circular segment of span \[12\text{ }cm\]with \[B\]as its...
Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.
Steps of construction: Define a boundary fragment, \[AB\text{ }=\text{ }7\text{ }cm.\] Draw a beam, \[AX\], making an intense point down ward with \[AB.\] Imprint the focuses\[{{A}_{1}},\text{...
To construct a triangle similar to a given △ABC with its sides 7/3 of the corresponding sides of △ABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect to BC. The points B1, B2, …., B7 are located at equal distances on BX, B3 is joined to C and then a line segment B6C‘ is drawn parallel to B3C where C‘ lies on BC produced. Finally, line segment A‘C‘ is drawn parallel to AC.
False Support: Allow us to attempt to build the figure as given in the inquiry. Steps of development, Define a boundary section \[BC.\] With \[B\text{ }and\text{ }C\]as focuses, draw two circular...
By geometrical construction, it is possible to divide a line segment in the ratio √3:(1/√3)
True Support: As per the inquiry, Ratio\[=\text{ }\surd 3\text{ }:\text{ }\left( \text{ }1/\surd 3 \right)\] On working on we get, \[\surd 3/\left( 1/\surd 3 \right)\text{ }=\text{ }\left( \surd...
Draw two concentric circles of radii \[\mathbf{3}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{5}\text{ }\mathbf{cm}.\]Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.
Steps of construction: Draw a circle with focus \[\mathbf{O}\]and radius \[\mathbf{3}\text{ }\mathbf{cm}.\] Draw one more circle with focus \[\mathbf{O}\]and radius \[\mathbf{5}\text{...
Draw a parallelogram \[\mathbf{ABCD}\]in which \[\mathbf{BC}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{AB}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{cm}\]and angle \[\mathbf{ABC}\text{ }=\text{ }\mathbf{60}{}^\circ ,\]divide it into triangles \[\mathbf{BCD}\]and \[\mathbf{ABD}\]by the diagonal \[\mathbf{BD}.\]Construct the triangle \[\mathbf{BD}~\mathbf{C}~\]similar to triangle \[\mathbf{BDC}\]with scale factor\[\mathbf{4}/\mathbf{3}\]. Draw the line segment \[\mathbf{D}\mathbf{A}~\]parallel to \[\mathbf{DA}\]where \[\mathbf{A}\] lies on extended side\[\mathbf{BA}\]. Is \[\mathbf{A}\mathbf{BC}\mathbf{D}\] a parallelogram?
Steps of construction: Define a boundary \[\mathbf{AB}=\mathbf{3}\text{ }\mathbf{cm}.\] Draw a beam \[\mathbf{BY}\]making an intense \[\angle \mathbf{ABY}=\mathbf{60}{}^\circ .\] With focus...
Two line segments \[\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AC}\]include an angle of \[\mathbf{60}{}^\circ \]where \[\mathbf{AB}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm}.\]Locate points \[\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{Q}\text{ }\mathbf{on}\text{ }\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AC},\]respectively such that \[\mathbf{AP}\text{ }=\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\text{ }\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AQ}\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }\mathbf{AC}.\] Join \[\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{Q}\]and measure the length \[\mathbf{PQ}.\]
Steps of construction: 1.Define a boundary portion \[AB\text{ }=\text{ }5\text{ }cm.\] Draw \[\angle BAZ\text{ }=\text{ }60{}^\circ .\] With focus \[A\]and sweep\[7\text{ }cm\], draw a...
To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}\text{ }:\text{ }\mathbf{6}\], draw a ray \[\mathbf{AX}\]such that \[\angle \mathbf{BAX}\] is an acute angle, then draw a ray \[\mathbf{BY}\]parallel to \[\mathbf{AX}\]and the points \[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\text{ }\ldots \]and \[{{\mathbf{B}}_{\mathbf{1}}},\text{ }{{\mathbf{B}}_{\mathbf{2}}},\text{ }{{\mathbf{B}}_{\mathbf{3}}},\text{ }\ldots \]are located at equal distances on ray \[\mathbf{AX}\]and\[\mathbf{BY}\], respectively. Then the points joined are \[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}~\] \[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{6}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}~\] \[~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{4}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}\] \[\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{4}}}\]
\[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}\] As per the inquiry, A line portion \[AB\]in the proportion \[5:7\] Along these...
To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{4}:\mathbf{7}\], a ray \[\mathbf{AX}\]is drawn first such that \[\mathbf{BAX}\]is an acute angle and then points \[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\ldots .\]are located at equal distances on the ray \[\mathbf{AX}\]and the point \[\mathbf{B}\]is joined to \[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{12}}}~\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{10}}}~\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{9}}}\]
SOLUTION:- \[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}\] As per the inquiry, A line section\[~AB\] in the proportion \[4:7\] Thus, \[A:B\text{ }=\text{ }4:7\] Presently, Draw a...
To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}:\mathbf{7}\], first a ray \[\mathbf{AX}\]is drawn so that \[\mathbf{BAX}\]is an acute angle and then at equal distances points are marked on the ray \[\mathbf{AX}\]such that the minimum number of these points is (A) \[\mathbf{8}\](B) \[\mathbf{10}\](C)\[~\mathbf{11}\] (D) \[\mathbf{12}\]
SOLUTION:- \[\left( D \right)\text{ }12\] As indicated by the inquiry, A line fragment \[AB\]in the proportion \[5:7\] In this way, \[A:B\text{ }=\text{ }5:7\] Presently, Draw a beam \[AX\]making an...
If PA and PB are tangents from an outside point P. such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.
Given, $AP=10cm$ $\angle APB={{60}^{\circ }}$ According to the figure We know that, A line drawn from centre to point from where external tangents are drawn, bisects the angle made by tangents at...
In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°, find ∠PRS
Given, $\angle TRQ={{30}^{\circ }}$ . At point R, OR ⊥ RQ. So, $\angle ORQ={{90}^{\circ }}$ $\Rightarrow \angle TRQ+\angle ORT={{90}^{\circ }}$ $\Rightarrow \angle ORT={{90}^{\circ...
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at centre.
We are considering a circle with centre ‘O’ with two parallel tangents through A & B at ends of diameter. M intersect the parallel tangents at P and Q Then, required to prove: $\angle...
If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of ∆APQ
Since AB and AC are the tangents from the same point A ∴AB=AC=5cm Similarly, BP=PX and XQ=QC Perimeter of \[\Delta APQ=AP+AQ+PQ\] \[=AP+AQ+(PX+XQ)\] \[=(AP+PX)+(AQ+XQ)\] \[=(AP+BP)+(AQ+QC)\]...
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.
Provided in question: Chord PQ is parallel to tangent at R.To prove: R bisects the arc PRQ. Proof: Since PQ || tangent at R. $\angle 1=\angle 2$ [alternate interior angles]$\angle 1=\angle 3$...
Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Suppose C1 and C2 are two circles with the same center O. And AC is a chord touching C1 at the point D let’s join OD.So, $OD\bot AC$$AD=DC=4cm$ [perpendicular line OD...
If the quadrilateral sides touch the circle, prove that sum of pair of opposite sides is equal to the sum of other pair.
Let’s Consider a quadrilateral ABCD touching circle with the centre O at points E, F, G and H as we can see in figure. We know that, In a circle with two points outside of it, the tangents drawn...
A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
Given, OP = $26cm$ PT = tangent length = $10cm$ To find: radius = OT =$?$ We know that, Radius and tangent are perpendicular at the point of contact, $\angle OTP={{90}^{\circ }}$ $$ So,...