Question

$\huge\blue{\boxed{\bf QueStion}}$

1.) If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 cm and ∠ADC = ∠BAC. Find the length of AC

2.) In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

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Answer 1

QUESTION NO 1 :-

• If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 cm and ∠ADC = ∠BAC. Find the length of AC.

Given :

• If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 cm and ∠ADC = ∠BAC.

To Find :-

• What is the length of AC.

Solution :-

Given :

• BC = 12 cm
• BD = 9 cm

As D is a point on the side BC then,

↦ BC = BD + CD

↦ 12 cm = 9 cm + CD

↦ 12 cm - 9 cm = CD

↦ 3 cm = CD

➠ CD = 3 cm

Now, in ∠ADC and ∠BAC,

⇒ ∠ADC = ∠BAC (Given)

Where,

⇒ ∠C = ∠C (Common angle)

➦ ∆BAC ~ ∆ADC (AA Similarly Criterion)

Now, we have to find the length of AC,

↦ $\sf \dfrac{BC}{AC} =\: \dfrac{AC}{CD}$

By doing cross multiplication we get,

↦ $\sf AC \times AC =\: BC \times CD$

↦ $\sf {(AC)}^{2} =\: 12 \times 3$

↦ $\sf {(AC)}^{2} =\: 36$

↦ $\sf AC =\: \sqrt{36}$

➦ $\sf\bold{\red{AC =\: 6\: cm}}$

$\therefore$ The length of AC is 6 cm.

________________________________________

QUESTION NO 2 :

• In an equilateral triangle, proved that three times the square of one side is equal to four times the square of one of its altitudes.

Given :

$\leadsto$ $\sf ABC\: is\: a\: equilateral\: triangle\:.$

$\leadsto$ $\sf AB =\: BC = AC$

$\leadsto$ $\sf AD \perp BC$

To Prove :

$\leadsto$ $\sf 3{AB}^{2} =\: 4{AD}^{2}$

Proof :

➲ In ∆ABC,

↦ AB = BC = AC

Again,

➲ In ∆ABD,

Now, by using Pythagoras theorem we get,

↦ $\sf {AB}^{2} =\: {BD}^{2} + {AD}^{2}$

↦ $\sf {AB}^{2} =\: {\bigg(\dfrac{BC}{2}\bigg)}^{2} + {AD}^{2}$

↦ $\sf {AB}^{2} =\: \dfrac{{BC}^{2}}{4} + {AD}^{2}$

↦ $\sf {AB}^{2} =\: \dfrac{{BC}^{2} + 4{AD}^{2}}{4}$

By doing cross multiplication we get,

↦ $\sf 4{AB}^{2} =\: {BC}^{2} + 4{AD}^{2}$

↦ $\sf 4{AB}^{2} - {BC}^{2} =\: 4{AD}^{2}$

↦ $\sf 4{AB}^{2} - {AB}^{2} =\: 4{AD}^{2}\: [\because AB = BC = AC]$

➠ $\sf\bold{\red{3{AB}^{2} =\: 4{AD}^{2}}}$

$\longmapsto$ Hence, Proved .

[Note : Please refer the attachment for the diagram.]

Answer 2

Answer:

QUESTION NO 1 :-

If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 cm and ∠ADC = ∠BAC. Find the length of AC.

Given :

If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 cm and ∠ADC = ∠BAC.

To Find :-

What is the length of AC.

Solution :-

Given :

BC = 12 cm

BD = 9 cm

As D is a point on the side BC then,

↦ BC = BD + CD

↦ 12 cm = 9 cm + CD

↦ 12 cm - 9 cm = CD

↦ 3 cm = CD

➠ CD = 3 cm

Now, in ∠ADC and ∠BAC,

⇒ ∠ADC = ∠BAC (Given)

Where,

⇒ ∠C = ∠C (Common angle)

➦ ∆BAC ~ ∆ADC (AA Similarly Criterion)

Now, we have to find the length of AC,

↦ \sf \dfrac{BC}{AC} =\: \dfrac{AC}{CD}

AC

BC

=

CD

AC

By doing cross multiplication we get,

↦ \sf AC \times AC =\: BC \times CDAC×AC=BC×CD

↦ \sf {(AC)}^{2} =\: 12 \times 3(AC)

2

=12×3

↦ \sf {(AC)}^{2} =\: 36(AC)

2

=36

↦ \sf AC =\: \sqrt{36}AC=

36

➦ \sf\bold{\red{AC =\: 6\: cm}}AC=6cm

\therefore∴ The length of AC is 6 cm.

________________________________________

QUESTION NO 2 :

In an equilateral triangle, proved that three times the square of one side is equal to four times the square of one of its altitudes.

Given :

\leadsto⇝ \sf ABC\: is\: a\: equilateral\: triangle\:.ABCisaequilateraltriangle.

\leadsto⇝ \sf AB =\: BC = ACAB=BC=AC

\leadsto⇝ \sf AD \perp BCAD⊥BC

To Prove :

\leadsto⇝ \sf 3{AB}^{2} =\: 4{AD}^{2}3AB

2

=4AD

2

Proof :

➲ In ∆ABC,

↦ AB = BC = AC

Again,

➲ In ∆ABD,

Now, by using Pythagoras theorem we get,

↦ \sf {AB}^{2} =\: {BD}^{2} + {AD}^{2}AB

2

=BD

2

+AD

2

↦ \sf {AB}^{2} =\: {\bigg(\dfrac{BC}{2}\bigg)}^{2} + {AD}^{2}AB

2

=(

2

BC

)

2

+AD

2

↦ \sf {AB}^{2} =\: \dfrac{{BC}^{2}}{4} + {AD}^{2}AB

2

=

4

BC

2

+AD

2

↦ \sf {AB}^{2} =\: \dfrac{{BC}^{2} + 4{AD}^{2}}{4}AB

2

=

4

BC

2

+4AD

2

By doing cross multiplication we get,

↦ \sf 4{AB}^{2} =\: {BC}^{2} + 4{AD}^{2}4AB

2

=BC

2

+4AD

2

↦ \sf 4{AB}^{2} - {BC}^{2} =\: 4{AD}^{2}4AB

2

−BC

2

=4AD

2

↦ \sf 4{AB}^{2} - {AB}^{2} =\: 4{AD}^{2}\: [\because AB = BC = AC]4AB

2

−AB

2

=4AD

2

[∵AB=BC=AC]

➠ \sf\bold{\red{3{AB}^{2} =\: 4{AD}^{2}}}3AB

2

=4AD

2

\longmapsto⟼ Hence, Proved .

[Note : Please refer the attachment for the diagram.]